Maharashtra Board Class 11 Maths Part 1 Chapter 3 Complex Numbers Miscellaneous Solutions

Question 1. Find the value of \( \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
Answer:
Given expression: \( \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
Taking out \( i^{10} \) as a common factor from the numerator:
\( = \frac{i^{10} (i^{582} + i^{580} + i^{578} + i^{576} + i^{574})}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
Cancelling the common terms in the numerator and denominator:
\( = i^{10} \)
Since \( i^2 = -1 \):
\( = (i^2)^5 \)
\( = (-1)^5 \)
\( = -1 \)
Thus, the simplified value of the given expression is \( -1 \).
In simple words: We can factor out \( i^{10} \) from the top part of the fraction. This leaves the exact same terms on the top and bottom, which cancel each other out, leaving us with just \( i^{10} \), which simplifies to -1.

🎯 Exam Tip: Always look for a common power of \( i \) to factor out from the numerator or denominator to simplify large expressions quickly instead of calculating each term individually.

Question 1. Simplify: \( \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
Answer:
\( \frac{i^{592} + i^{590} + i^{588} + i^{586} + i^{584}}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
\( = \frac{i^{10}(i^{582} + i^{580} + i^{578} + i^{576} + i^{574})}{i^{582} + i^{580} + i^{578} + i^{576} + i^{574}} \)
\( = i^{10} \)
\( = (i^4)^2 \cdot i^2 \)
\( = (1)^2 (-1) \)
\( = -1 \)
This elegant cancellation simplifies the expression significantly.
In simple words: We can factor out \( i^{10} \) from all terms in the numerator. This makes the numerator and denominator identical except for the \( i^{10} \) factor, allowing them to cancel out and simplify the expression to \(-1\).

🎯 Exam Tip: When dealing with large consecutive powers of \( i \), look for common factors to simplify the fraction quickly before calculating individual powers.

Question 2. Find the value of \( \sqrt{-3} \times \sqrt{-6} \).
Answer:
\( \sqrt{-3} \times \sqrt{-6} = \sqrt{3} \times \sqrt{-1} \times \sqrt{6} \times \sqrt{-1} \)
\( = \sqrt{3}i \times \sqrt{6}i \)
\( = \sqrt{18}i^2 \)
\( = -3\sqrt{2} \quad [\because i^2 = -1] \)
This shows how negative roots lead to real coefficients when multiplied.
In simple words: First, convert the square roots of negative numbers into imaginary numbers using \( i \). Then multiply them together, remembering that \( i^2 = -1 \).

🎯 Exam Tip: Never multiply negative numbers directly under square roots like \( \sqrt{-3} \times \sqrt{-6} = \sqrt{18} \). Always convert them to the \( i \) form first to avoid sign errors.

Question 3. Simplify the following and express in the form a + ib.
(i) \( 3 + \sqrt{-64} \)
(ii) \( (2i^3)^2 \)
(iii) \( (2 + 3i) (1 - 4i) \)
(iv) \( \frac{5}{2}i(-4 - 3i) \)
(v) \( (1 + 3i)^2 (3 + i) \)
(vi) \( \frac{4+3i}{1-i} \)
(vii) \( (1 + \frac{2}{i}) (3 + \frac{4}{i}) (5 + i)^{-1} \)
(viii) \( \frac{\sqrt{5}+\sqrt{3}i}{\sqrt{5}-\sqrt{3}i} \)
(ix) \( \frac{3i^5+2i^7+i^9}{i^6+2i^8+3i^{18}} \)
(x) \( \frac{5+7i}{4+3i} + \frac{5+7i}{4-3i} \)
Answer:
(i) \( 3 + \sqrt{-64} \)
\( = 3 + \sqrt{64} \cdot \sqrt{-1} \)
\( = 3 + 8i \)

(ii) \( (2i^3)^2 \)
\( = 4i^6 \)
\( = 4(i^2)^3 \)
\( = 4(-1)^3 \)
\( = -4 \)
\( = -4 + 0i \)

(iii) \( (2 + 3i) (1 - 4i) \)
\( = 2(1 - 4i) + 3i(1 - 4i) \)
\( = 2 - 8i + 3i - 12i^2 \)
\( = 2 - 5i - 12(-1) \)
\( = 2 - 5i + 12 \)
\( = 14 - 5i \)

(iv) \( \frac{5}{2}i(-4 - 3i) \)
\( = \frac{5}{2}i(-4) - \frac{5}{2}i(3i) \)
\( = -10i - \frac{15}{2}i^2 \)
\( = -10i - \frac{15}{2}(-1) \)
\( = \frac{15}{2} - 10i \)

(v) \( (1 + 3i)^2 (3 + i) \)
\( = (1 + 6i + 9i^2)(3 + i) \)
\( = (1 + 6i - 9)(3 + i) \)
\( = (-8 + 6i)(3 + i) \)
\( = -8(3 + i) + 6i(3 + i) \)
\( = -24 - 8i + 18i + 6i^2 \)
\( = -24 + 10i - 6 \)
\( = -30 + 10i \)

(vi) \( \frac{4+3i}{1-i} \)
\( = \frac{4+3i}{1-i} \times \frac{1+i}{1+i} \)
\( = \frac{4(1+i) + 3i(1+i)}{1^2 - i^2} \)
\( = \frac{4 + 4i + 3i + 3i^2}{1 - (-1)} \)
\( = \frac{4 + 7i - 3}{2} \)
\( = \frac{1 + 7i}{2} \)
\( = \frac{1}{2} + \frac{7}{2}i \)

(vii) \( (1 + \frac{2}{i}) (3 + \frac{4}{i}) (5 + i)^{-1} \)
Since \( \frac{1}{i} = -i \):
\( = (1 - 2i) (3 - 4i) \frac{1}{5+i} \)
\( = (3 - 4i - 6i + 8i^2) \frac{1}{5+i} \)
\( = (3 - 10i - 8) \frac{1}{5+i} \)
\( = \frac{-5 - 10i}{5+i} \)
\( = \frac{-5(1 + 2i)}{5+i} \times \frac{5-i}{5-i} \)
\( = \frac{-5(5 - i + 10i - 2i^2)}{5^2 - i^2} \)
\( = \frac{-5(5 + 9i + 2)}{25 - (-1)} \)
\( = \frac{-5(7 + 9i)}{26} \)
\( = -\frac{35}{26} - \frac{45}{26}i \)

(viii) \( \frac{\sqrt{5}+\sqrt{3}i}{\sqrt{5}-\sqrt{3}i} \)
\( = \frac{\sqrt{5}+\sqrt{3}i}{\sqrt{5}-\sqrt{3}i} \times \frac{\sqrt{5}+\sqrt{3}i}{\sqrt{5}+\sqrt{3}i} \)
\( = \frac{(\sqrt{5}+\sqrt{3}i)^2}{(\sqrt{5})^2 - (\sqrt{3}i)^2} \)
\( = \frac{5 + 2\sqrt{15}i + 3i^2}{5 - 3i^2} \)
\( = \frac{5 + 2\sqrt{15}i - 3}{5 + 3} \)
\( = \frac{2 + 2\sqrt{15}i}{8} \)
\( = \frac{1}{4} + \frac{\sqrt{15}}{4}i \)

(ix) \( \frac{3i^5+2i^7+i^9}{i^6+2i^8+3i^{18}} \)
Since \( i^5 = i \), \( i^7 = -i \), \( i^9 = i \), \( i^6 = -1 \), \( i^8 = 1 \), \( i^{18} = -1 \):
\( = \frac{3(i) + 2(-i) + i}{-1 + 2(1) + 3(-1)} \)
\( = \frac{3i - 2i + i}{-1 + 2 - 3} \)
\( = \frac{2i}{-2} \)
\( = -i \)
\( = 0 - i \)

(x) \( \frac{5+7i}{4+3i} + \frac{5+7i}{4-3i} \)
\( = (5+7i) \left[ \frac{1}{4+3i} + \frac{1}{4-3i} \right] \)
\( = (5+7i) \left[ \frac{(4-3i) + (4+3i)}{(4+3i)(4-3i)} \right] \)
\( = (5+7i) \left[ \frac{8}{4^2 - (3i)^2} \right] \)
\( = (5+7i) \left[ \frac{8}{16 + 9} \right] \)
\( = (5+7i) \left[ \frac{8}{25} \right] \)
\( = \frac{40 + 56i}{25} \)
\( = \frac{40}{25} + \frac{56}{25}i \)
\( = \frac{8}{5} + \frac{56}{25}i \)
Using these standard algebraic identities helps systematically reduce complex fractions.
In simple words: To express any complex expression in the standard form \( a + ib \), simplify the powers of \( i \), perform algebraic operations, and rationalize denominators by multiplying by their complex conjugates.

🎯 Exam Tip: Always write the final answer clearly in the form \( a + ib \), even if \( a \) or \( b \) is zero (e.g., write \( -4 \) as \( -4 + 0i \)).

Question. Simplify the following and express in the form of \( a + ib \):
(ii) \( (2i^3)^2 \)
(iii) \( (2 + 3i)(1 - 4i) \)
(iv) \( \frac{5}{2}i(-4 - 3i) \)
(v) \( (1 + 3i)^2(3 + i) \)
Answer:
(ii) \( (2i^3)^2 \)
\( = 4i^6 \)
\( = 4(i^2)^3 \)
\( = 4(-1)^3 \quad [\because i^2 = -1] \)
\( = -4 \)
\( = -4 + 0i \)

(iii) \( (2 + 3i)(1 - 4i) \)
\( = 2 - 8i + 3i - 12i^2 \)
\( = 2 - 5i - 12(-1) \quad [\because i^2 = -1] \)
\( = 14 - 5i \)

(iv) \( \frac{5}{2}i(-4 - 3i) \)
\( = \frac{5}{2}(-4i - 3i^2) \)
\( = \frac{5}{2}[-4i - 3(-1)] \quad [\because i^2 = -1] \)
\( = \frac{5}{2}(3 - 4i) \)
\( = \frac{15}{2} - 10i \)

(v) \( (1 + 3i)^2(3 + i) \)
\( = (1 + 6i + 9i^2)(3 + i) \)
\( = (1 + 6i - 9)(3 + i) \quad [\times i^2 = -1] \)
\( = (-8 + 6i)(3 + i) \)
\( = -24 - 8i + 18i + 6i^2 \)
\( = -24 + 10i + 6(-1) \)
\( = -24 + 10i - 6 \)
\( = -30 + 10i \)
These steps systematically resolve the powers of the imaginary unit to yield standard complex numbers.
In simple words: To simplify these expressions, we multiply the terms normally and replace \( i^2 \) with \( -1 \) whenever it appears. Finally, we group the real numbers together and the imaginary numbers together to get the final answer.

🎯 Exam Tip: Always remember that \( i^2 = -1 \). Carefully substitute this value and group the real and imaginary parts separately to avoid sign errors.

Question. Simplify the following complex numbers and express them in the form \( a + ib \):
vi. \( \frac{4+3i}{1-i} \)
vii. \( \left(1 + \frac{2}{i}\right)\left(3 + \frac{4}{i}\right)(5 + i)^{-1} \)
Answer:
vi.
\( \frac{4+3i}{1-i} = \frac{(4+3i)(1+i)}{(1-i)(1+i)} \)
\( = \frac{4 + 4i + 3i + 3i^2}{1 - i^2} \)
\( = \frac{4 + 7i + 3(-1)}{1 - (-1)} \quad \dots [\because i^2 = -1] \)
\( = \frac{1 + 7i}{2} = \frac{1}{2} + \frac{7}{2}i \)

vii.
\( \left(1 + \frac{2}{i}\right)\left(3 + \frac{4}{i}\right)(5 + i)^{-1} \)
\( = \frac{(i+2)}{i} \cdot \frac{(3i+4)}{i} \cdot \frac{1}{5+i} \)
\( = \frac{3i^2 + 4i + 6i + 8}{i^2(5+i)} \)
\( = \frac{-3 + 10i + 8}{-1(5+i)} \quad \dots [\because i^2 = -1] \)
\( = \frac{5+10i}{-(5+i)} \)
\( = \frac{(5+10i)(5-i)}{-(5+i)(5-i)} \)
\( = \frac{25 - 5i + 50i - 10i^2}{-(25 - i^2)} \)
\( = \frac{25 + 45i - 10(-1)}{-[25 - (-1)]} \)
\( = \frac{35 + 45i}{-26} \)
\( = -\frac{35}{26} - \frac{45}{26}i \)
In simple words: To simplify these expressions, we perform basic algebra while replacing \( i^2 \) with \( -1 \). If there is an \( i \) in the denominator, we multiply both the top and bottom by its conjugate to clear it out.

🎯 Exam Tip: Be extremely careful with negative signs when substituting \( i^2 = -1 \), especially when simplifying denominators like \( -(25 - i^2) \).

Question viii. Simplify \( \frac{\sqrt{5} + \sqrt{3}i}{\sqrt{5} - \sqrt{3}i} \)
Answer:
\( \frac{\sqrt{5} + \sqrt{3}i}{\sqrt{5} - \sqrt{3}i} \)
\( = \frac{(\sqrt{5} + \sqrt{3}i)(\sqrt{5} + \sqrt{3}i)}{(\sqrt{5} - \sqrt{3}i)(\sqrt{5} + \sqrt{3}i)} \)
\( = \frac{5 + 2\sqrt{15}i + 3i^2}{5 - 3i^2} \)
\( = \frac{5 + 2\sqrt{15}i + 3(-1)}{5 - 3(-1)} \)     ...[\( \because i^2 = -1 \)]
\( = \frac{2 + 2\sqrt{15}i}{8} = \frac{1 + \sqrt{15}i}{4} \)
\( = \frac{1}{4} + \frac{\sqrt{15}}{4}i \)
This complex number is now successfully written in its standard algebraic form.
In simple words: To simplify a fraction with an imaginary number at the bottom, we multiply both the top and bottom by its opposite partner (conjugate) to clear the square root and imaginary parts from the denominator.

🎯 Exam Tip: Always remember to substitute \( i^2 = -1 \) and simplify the real terms together before writing the final answer in \( a + ib \) form.

Question ix. Simplify \( \frac{3i^5 + 2i^7 + i^9}{i^6 + 2i^8 + 3i^{18}} \)
Answer:
\( \frac{3i^5 + 2i^7 + i^9}{i^6 + 2i^8 + 3i^{18}} = \frac{3(i^4 \cdot i) + 2(i^4 \cdot i^3) + (i^4)^2 \cdot i}{i^4 \cdot i^2 + 2(i^4)^2 + 3(i^2)^9} \)
\( = \frac{3(1) \cdot i + 2(1)(-i) + (1)^2 \cdot i}{(1)(-1) + 2(1)^2 + 3(-1)^9} \)     ...[\( \because i^2 = -1, i^3 = -i, i^4 = 1 \)]
\( = \frac{3i - 2i + i}{-1 + 2 - 3} \)
\( = \frac{2i}{-2} \)
\( = -i = 0 - i \)
This reduces the complex expression to a very simple imaginary unit.
In simple words: We can simplify high powers of \( i \) by breaking them down using \( i^4 = 1 \) and \( i^2 = -1 \), which makes the big expression much easier to solve.

🎯 Exam Tip: Express higher powers of \( i \) in terms of \( i^4 \) because \( i^4 = 1 \), which drastically simplifies calculations and prevents arithmetic errors.

Question x. Simplify \( \frac{5+7i}{4+3i} + \frac{5+7i}{4-3i} \)
Answer:
\( \frac{5+7i}{4+3i} + \frac{5+7i}{4-3i} \)
\( = (5 + 7i) \left[ \frac{1}{4+3i} + \frac{1}{4-3i} \right] \)
\( = (5 + 7i) \left[ \frac{4-3i + 4+3i}{(4+3i)(4-3i)} \right] \)
\( = (5 + 7i) \left[ \frac{8}{16 - 9i^2} \right] \)
\( = (5 + 7i) \left[ \frac{8}{16 - 9(-1)} \right] \)     ...[\( \because i^2 = -1 \)]
\( = \frac{8(5+7i)}{25} = \frac{40 + 56i}{25} \)
\( = \frac{40}{25} + \frac{56}{25}i = \frac{8}{5} + \frac{56}{25}i \)
This shows how factoring out common terms can make complex fraction addition much cleaner.
In simple words: Instead of simplifying both fractions separately, we can pull out the common top part first, add the simpler fractions inside, and then multiply at the end.

🎯 Exam Tip: Factoring out common terms like \( (5+7i) \) first saves a lot of calculation time and reduces the chance of making algebraic mistakes.

Question 4. Solve the following equations for \( x, y \in \mathbb{R} \):
(i) \( (4 - 5i)x + (2 + 3i)y = 10 - 7i \)
Answer:
Given equation:
\( (4 - 5i)x + (2 + 3i)y = 10 - 7i \)
Expanding the terms:
\( 4x - 5xi + 2y + 3yi = 10 - 7i \)
Grouping real and imaginary parts:
\( (4x + 2y) + (-5x + 3y)i = 10 - 7i \)
Equating the real and imaginary parts on both sides:
\( 4x + 2y = 10 \)     ... (1)
\( -5x + 3y = -7 \)     ... (2)
Dividing equation (1) by 2:

\( \implies 2x + y = 5 \)

\( \implies y = 5 - 2x \)     ... (3)
Substituting equation (3) in equation (2):
\( -5x + 3(5 - 2x) = -7 \)
\( -5x + 15 - 6x = -7 \)
\( -11x + 15 = -7 \)
\( -11x = -22 \)

\( \implies x = 2 \)
Substituting \( x = 2 \) in equation (3):
\( y = 5 - 2(2) \)

\( \implies y = 1 \)
Thus, the values are \( x = 2 \) and \( y = 1 \).
In simple words: To solve this, we separate the real numbers from the imaginary numbers (the ones with \( i \)) on both sides, set up two simple equations, and solve them to find \( x \) and \( y \).

🎯 Exam Tip: When equating complex numbers, always clearly state that you are "equating real and imaginary parts" to show your step-by-step logical reasoning to the examiner.

Question 1. Find the values of \( x \) and \( y \) which satisfy the equation: \( (4 - 5i)x + (2 + 3i)y = 10 - 7i \)
Answer:
Given equation: \( (4 - 5i)x + (2 + 3i)y = 10 - 7i \)
\( \implies (4x + 2y) + (3y - 5x)i = 10 - 7i \)
Equating real and imaginary parts, we get:
\( 4x + 2y = 10 \)
i.e., \( 2x + y = 5 \) ......(i)
and \( 3y - 5x = -7 \) ......(ii)
Equation (i) \( \times 3 \) \( - \) equation (ii) gives:
\( 11x = 22 \)
\( \implies x = 2 \)
Putting \( x = 2 \) in (i), we get:
\( 2(2) + y = 5 \)
\( \implies y = 1 \)
\( \implies x = 2 \) and \( y = 1 \)
In simple words: To find \( x \) and \( y \), we group the real numbers together and the imaginary numbers (with \( i \)) together on both sides. Then, we set them equal to each other to solve like normal simultaneous equations.

🎯 Exam Tip: Always clearly state the step where you equate the real and imaginary parts, as examiners look for this key step to award partial marks.

Question 2. Find the values of \( x \) and \( y \) which satisfy the equation: \( (1 - 3i)x + (2 + 5i)y = 7 + i \)
Answer:
Given equation: \( (1 - 3i)x + (2 + 5i)y = 7 + i \)
\( \implies (x + 2y) + (-3x + 5y)i = 7 + i \)
Equating real and imaginary parts, we get:
\( x + 2y = 7 \) ......(i)
and \( -3x + 5y = 1 \) ......(ii)
Equation (i) \( \times 3 \) + equation (ii) gives:
\( 11y = 22 \)
\( \implies y = 2 \)
Putting \( y = 2 \) in (i), we get:
\( x + 2(2) = 7 \)
\( \implies x = 3 \)
\( \implies x = 3 \) and \( y = 2 \)
In simple words: We separate the parts with \( i \) from the parts without \( i \) on both sides of the equals sign. This gives us two simple equations that we can solve to find the values of \( x \) and \( y \).

🎯 Exam Tip: Double-check your signs when multiplying equations to eliminate variables, as a simple sign error can lead to incorrect values for \( x \) and \( y \).

Question 3. Find the values of \( x \) and \( y \) which satisfy the equation: \( \frac{x+iy}{2+3i} = 7 - i \)
Answer:
Given equation: \( \frac{x+iy}{2+3i} = 7 - i \)
\( \implies x + iy = (7 - i)(2 + 3i) \)
\( \implies x + iy = 14 + 21i - 2i - 3i^2 \)
\( \implies x + iy = 14 + 19i - 3(-1) \) .....[\( \because i^2 = -1 \)]
\( \implies x + iy = 14 + 19i + 3 \)
\( \implies x + iy = 17 + 19i \)
Equating real and imaginary parts, we get:
\( x = 17 \) and \( y = 19 \)
In simple words: We multiply both sides by the denominator to get \( x + iy \) by itself. Then we multiply the complex numbers on the right side and use \( i^2 = -1 \) to simplify and find \( x \) and \( y \).

🎯 Exam Tip: Remember that \( i^2 = -1 \). Replacing \( i^2 \) with \( -1 \) is the most common place where students make sign errors, so write down this step explicitly.

Question 4. Find the values of \( x \) and \( y \) which satisfy the following equations:
(iv) \( (x + iy)(5 + 6i) = 2 + 3i \)
(v) \( 2x + i^9 y(2 + i) = x i^7 + 10 i^{16} \)
Answer:
(iv) \( (x + iy)(5 + 6i) = 2 + 3i \)
\( \implies x + iy = \frac{2 + 3i}{5 + 6i} \)
\( \implies x + iy = \frac{(2 + 3i)(5 - 6i)}{(5 + 6i)(5 - 6i)} \)
\( \implies x + iy = \frac{10 - 12i + 15i - 18i^2}{25 - 36i^2} \)
\( \implies x + iy = \frac{10 + 3i - 18(-1)}{25 - 36(-1)} \) ... [\( \because i^2 = -1 \)]
\( \implies x + iy = \frac{10 + 3i + 18}{25 + 36} \)
\( \implies x + iy = \frac{28 + 3i}{61} \)
\( \implies x + iy = \frac{28}{61} + \frac{3}{61}i \)
Equating real and imaginary parts, we get:
\( x = \frac{28}{61} \) and \( y = \frac{3}{61} \)

(v) \( 2x + i^9 y(2 + i) = x i^7 + 10 i^{16} \)
\( \implies 2x + (i^4)^2 \cdot i \cdot y(2 + i) = x (i^2)^3 \cdot i + 10 \cdot (i^4)^4 \)
\( \implies 2x + (1)^2 \cdot iy(2 + i) = x (-1)^3 \cdot i + 10(1)^4 \) ... [\( \because i^2 = -1, i^4 = 1 \)]
\( \implies 2x + 2yi + yi^2 = -xi + 10 \)
\( \implies 2x + 2yi - y + xi = 10 \)
\( \implies (2x - y) + (x + 2y)i = 10 + 0i \)
Equating real and imaginary parts, we get:
\( 2x - y = 10 \) ... (i)
and \( x + 2y = 0 \) ... (ii)
Equation (i) \( \times 2 + \) equation (ii) gives:
\( 5x = 20 \)
\( \implies x = 4 \)
Putting \( x = 4 \) in equation (i), we get:
\( 2(4) - y = 10 \)
\( \implies 8 - y = 10 \)
\( \implies y = 8 - 10 \)
\( \implies y = -2 \)
\( \implies x = 4 \) and \( y = -2 \)
In simple words: To find x and y, we simplify the complex equations by replacing powers of i (like \( i^2 = -1 \)) and then separate the real parts and imaginary parts to solve them like normal simultaneous equations.

🎯 Exam Tip: Always simplify higher powers of \( i \) first using \( i^4 = 1 \) and \( i^2 = -1 \) to make the equations much easier to solve.

Question 5. Find the value of:
(i) \( x^3 + 2x^2 - 3x + 21 \), if \( x = 1 + 2i \)
Answer:
Given, \( x = 1 + 2i \)
\( \implies x - 1 = 2i \)
Squaring both sides, we get:
\( (x - 1)^2 = (2i)^2 \)
\( \implies x^2 - 2x + 1 = 4i^2 \)
\( \implies x^2 - 2x + 1 = -4 \) ... [\( \because i^2 = -1 \)]
\( \implies x^2 - 2x + 5 = 0 \)
Now, we express the given polynomial \( x^3 + 2x^2 - 3x + 21 \) in terms of \( x^2 - 2x + 5 \):
By performing polynomial division or algebraic adjustment:
\( x^3 + 2x^2 - 3x + 21 = x(x^2 - 2x + 5) + 4x^2 - 8x + 21 \)
\( = x(x^2 - 2x + 5) + 4(x^2 - 2x + 5) + 1 \)
Substituting \( x^2 - 2x + 5 = 0 \):
\( = x(0) + 4(0) + 1 \)
\( = 1 \)
Therefore, the value of \( x^3 + 2x^2 - 3x + 21 \) is 1.
In simple words: Instead of directly putting the complex value of x into the long expression, we find a simpler quadratic equation that equals zero and use it to simplify the larger expression.

🎯 Exam Tip: When finding the value of a high-degree polynomial for a complex number, always form a quadratic equation equal to zero first, then divide the polynomial by this quadratic to find the remainder quickly.

Question 1. Find the value of the polynomial expressions:
(i) \( x^3 + 2x^2 - 3x + 21 \), if \( x = 1 + 2i \)
(ii) \( x^3 - 5x^2 + 4x + 8 \), if \( x = \frac{10}{3-i} \)
(iii) \( x^3 - 3x^2 + 19x - 20 \), if \( x = 1 - 4i \)
Answer:
(i) Given \( x = 1 + 2i \)
\( x - 1 = 2i \)
Squaring both sides:
\( \implies (x - 1)^2 = (2i)^2 \)
\( \implies x^2 - 2x + 1 = 4i^2 \)
\( \implies x^2 - 2x + 1 = -4 \)    ...[\( \because i^2 = -1 \)]
\( \implies x^2 - 2x + 5 = 0 \)    ...(1)
Now, dividing \( x^3 + 2x^2 - 3x + 21 \) by \( x^2 - 2x + 5 \):
\( x^3 + 2x^2 - 3x + 21 = (x^2 - 2x + 5)(x + 4) + 1 \)
Substituting the value from (1):
\( \implies x^3 + 2x^2 - 3x + 21 = 0 \cdot (x + 4) + 1 \)
\( \implies x^3 + 2x^2 - 3x + 21 = 1 \)

(ii) Given \( x = \frac{10}{3-i} \)
Rationalizing the denominator:
\( x = \frac{10(3+i)}{(3-i)(3+i)} \)
\( \implies x = \frac{10(3+i)}{9 - i^2} \)
\( \implies x = \frac{10(3+i)}{9 - (-1)} \)
\( \implies x = \frac{10(3+i)}{10} \)
\( \implies x = 3 + i \)
\( \implies x - 3 = i \)
Squaring both sides:
\( \implies (x - 3)^2 = i^2 \)
\( \implies x^2 - 6x + 9 = -1 \)
\( \implies x^2 - 6x + 10 = 0 \)    ...(2)
Now, dividing \( x^3 - 5x^2 + 4x + 8 \) by \( x^2 - 6x + 10 \):
\( x^3 - 5x^2 + 4x + 8 = (x^2 - 6x + 10)(x + 1) - 2 \)
Substituting the value from (2):
\( \implies x^3 - 5x^2 + 4x + 8 = 0 \cdot (x + 1) - 2 \)
\( \implies x^3 - 5x^2 + 4x + 8 = -2 \)

(iii) Given \( x = 1 - 4i \)
\( x - 1 = -4i \)
Squaring both sides:
\( \implies (x - 1)^2 = (-4i)^2 \)
\( \implies x^2 - 2x + 1 = 16i^2 \)
\( \implies x^2 - 2x + 1 = -16 \)    ...[\( \because i^2 = -1 \)]
\( \implies x^2 - 2x + 17 = 0 \)    ...(3)
Now, dividing \( x^3 - 3x^2 + 19x - 20 \) by \( x^2 - 2x + 17 \):
\( x^3 - 3x^2 + 19x - 20 = (x^2 - 2x + 17)(x - 1) - 3 \)
Substituting the value from (3):
\( \implies x^3 - 3x^2 + 19x - 20 = 0 \cdot (x - 1) - 3 \)
\( \implies x^3 - 3x^2 + 19x - 20 = -3 \)
In simple words: To find the value of these long expressions, we first simplify the given value of x to form a quadratic equation equal to zero. Then, we divide the main expression by this quadratic equation so that the large part becomes zero, leaving us with a very simple final number.

🎯 Exam Tip: When simplifying complex numbers, always eliminate the imaginary unit \( i \) by isolating it on one side and squaring both sides to form a quadratic equation equal to zero.

Question. Solve the following:
(ii) Find the value of \( x^3 - 5x^2 + 4x + 8 \) if \( x = \frac{10}{3-i} \).
(iii) Find the value of \( x^2 - 2x + 17 \) if \( x = 1 - 4i \).
Answer:
(ii) Given,
\( x = \frac{10}{3-i} \)
\( \therefore x = \frac{10(3+i)}{(3-i)(3+i)} \)
\( = \frac{10(3+i)}{9-i^2} \)
\( = \frac{10(3+i)}{9-(-1)} \) ... [\( \because i^2 = -1 \)]
\( = \frac{10(3+i)}{10} \)
\( \therefore x = 3 + i \)
\( \therefore x - 3 = i \)
\( \dots (x - 3)^2 = i^2 \)
\( \therefore x^2 - 6x + 9 = -1 \) ... [\( \because i^2 = -1 \)]
\( \dots x^2 - 6x + 10 = 0 \) ......(i)

By dividing \( x^3 - 5x^2 + 4x + 8 \) by \( x^2 - 6x + 10 \), we can express the polynomial as:
\( x^3 - 5x^2 + 4x + 8 = (x^2 - 6x + 10)(x + 1) - 2 \)
\( = 0 \cdot (x + 1) - 2 \) ......[From (i)]
\( = 0 - 2 \)
\( \therefore x^3 - 5x^2 + 4x + 8 = -2 \)

(iii) Given,
\( x = 1 - 4i \)
\( \therefore x - 1 = -4i \)
\( \therefore (x - 1)^2 = (-4i)^2 \)
\( \therefore (x - 1)^2 = 16i^2 \)
\( \therefore x^2 - 2x + 1 = -16 \) ......[\( \because i^2 = -1 \)]
\( \therefore x^2 - 2x + 17 = 0 \) ......(i)
This method of using polynomial division helps us evaluate high-degree expressions easily without directly substituting complex values.
In simple words: To find the value of a complex expression, we first simplify the given value of \( x \) to form a quadratic equation equal to zero. Then, we divide the larger expression by this quadratic equation to easily find the final numerical value.

🎯 Exam Tip: Always show the substitution \( i^2 = -1 \) clearly in brackets, as examiners look for this key step when grading complex number simplifications.

Question 6. Find the square roots of:
(i) \( -16 + 30i \)
(ii) \( 15 - 8i \)
(iii) \( 2 + 2\sqrt{3}i \)
(iv) \( 18i \)
(v) \( 3 - 4i \)
(vi) \( 6 + 8i \)
Answer:
(i) \( -16 + 30i \)
Let \( \sqrt{-16 + 30i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( -16 + 30i = (a + bi)^2 \)
\( \implies -16 + 30i = a^2 + b^2i^2 + 2abi \)
\( \implies -16 + 30i = (a^2 - b^2) + 2abi \) [Since \( i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = -16 \) and \( 2ab = 30 \)
\( \implies a^2 - b^2 = -16 \) and \( b = \frac{15}{a} \)
Substituting the value of \( b \) in the first equation:
\( a^2 - \left(\frac{15}{a}\right)^2 = -16 \)
\( \implies a^2 - \frac{225}{a^2} = -16 \)
\( \implies a^4 - 225 = -16a^2 \)
\( \implies a^4 + 16a^2 - 225 = 0 \)
\( \implies (a^2 + 25)(a^2 - 9) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \geq 0 \), so \( a^2 \neq -25 \).
\( \implies a^2 = 9 \)
\( \implies a = \pm 3 \)
When \( a = 3 \), \( b = \frac{15}{3} = 5 \).
When \( a = -3 \), \( b = \frac{15}{-3} = -5 \).
Thus, the square roots of \( -16 + 30i \) are \( \pm(3 + 5i) \). This algebraic method is highly reliable for finding complex roots.

(ii) \( 15 - 8i \)
Let \( \sqrt{15 - 8i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 15 - 8i = (a + bi)^2 \)
\( \implies 15 - 8i = (a^2 - b^2) + 2abi \)
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 15 \) and \( 2ab = -8 \)
\( \implies a^2 - b^2 = 15 \) and \( b = -\frac{4}{a} \)
Substituting the value of \( b \) in the first equation:
\( a^2 - \left(-\frac{4}{a}\right)^2 = 15 \)
\( \implies a^2 - \frac{16}{a^2} = 15 \)
\( \implies a^4 - 16 = 15a^2 \)
\( \implies a^4 - 15a^2 - 16 = 0 \)
\( \implies (a^2 - 16)(a^2 + 1) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \neq -1 \).
\( \implies a^2 = 16 \)
\( \implies a = \pm 4 \)
When \( a = 4 \), \( b = -\frac{4}{4} = -1 \).
When \( a = -4 \), \( b = -\frac{4}{-4} = 1 \).
Thus, the square roots of \( 15 - 8i \) are \( \pm(4 - i) \).

(iii) \( 2 + 2\sqrt{3}i \)
Let \( \sqrt{2 + 2\sqrt{3}i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 2 + 2\sqrt{3}i = (a^2 - b^2) + 2abi \)
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 2 \) and \( 2ab = 2\sqrt{3} \)
\( \implies a^2 - b^2 = 2 \) and \( b = \frac{\sqrt{3}}{a} \)
Substituting the value of \( b \) in the first equation:
\( a^2 - \left(\frac{\sqrt{3}}{a}\right)^2 = 2 \)
\( \implies a^2 - \frac{3}{a^2} = 2 \)
\( \implies a^4 - 3 = 2a^2 \)
\( \implies a^4 - 2a^2 - 3 = 0 \)
\( \implies (a^2 - 3)(a^2 + 1) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \neq -1 \).
\( \implies a^2 = 3 \)
\( \implies a = \pm \sqrt{3} \)
When \( a = \sqrt{3} \), \( b = \frac{\sqrt{3}}{\sqrt{3}} = 1 \).
When \( a = -\sqrt{3} \), \( b = \frac{\sqrt{3}}{-\sqrt{3}} = -1 \).
Thus, the square roots of \( 2 + 2\sqrt{3}i \) are \( \pm(\sqrt{3} + i) \).

(iv) \( 18i \)
Let \( \sqrt{18i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 18i = (a^2 - b^2) + 2abi \)
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 0 \) and \( 2ab = 18 \)
\( \implies a^2 = b^2 \) and \( ab = 9 \)
Since \( ab = 9 > 0 \), \( a \) and \( b \) must have the same sign, so \( a = b \).
Substituting \( a = b \) in \( ab = 9 \):
\( a^2 = 9 \)
\( \implies a = \pm 3 \)
When \( a = 3 \), \( b = 3 \).
When \( a = -3 \), \( b = -3 \).
Thus, the square roots of \( 18i \) are \( \pm(3 + 3i) \).

(v) \( 3 - 4i \)
Let \( \sqrt{3 - 4i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 3 - 4i = (a^2 - b^2) + 2abi \)
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 3 \) and \( 2ab = -4 \)
\( \implies a^2 - b^2 = 3 \) and \( b = -\frac{2}{a} \)
Substituting the value of \( b \) in the first equation:
\( a^2 - \left(-\frac{2}{a}\right)^2 = 3 \)
\( \implies a^2 - \frac{4}{a^2} = 3 \)
\( \implies a^4 - 4 = 3a^2 \)
\( \implies a^4 - 3a^2 - 4 = 0 \)
\( \implies (a^2 - 4)(a^2 + 1) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \neq -1 \).
\( \implies a^2 = 4 \)
\( \implies a = \pm 2 \)
When \( a = 2 \), \( b = -\frac{2}{2} = -1 \).
When \( a = -2 \), \( b = -\frac{2}{-2} = 1 \).
Thus, the square roots of \( 3 - 4i \) are \( \pm(2 - i) \).

(vi) \( 6 + 8i \)
Let \( \sqrt{6 + 8i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 6 + 8i = (a^2 - b^2) + 2abi \)
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 6 \) and \( 2ab = 8 \)
\( \implies a^2 - b^2 = 6 \) and \( b = \frac{4}{a} \)
Substituting the value of \( b \) in the first equation:
\( a^2 - \left(\frac{4}{a}\right)^2 = 6 \)
\( \implies a^2 - \frac{16}{a^2} = 6 \)
\( \implies a^4 - 16 = 6a^2 \)
\( \implies a^4 - 6a^2 - 16 = 0 \)
\( \implies (a^2 - 8)(a^2 + 2) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \neq -2 \).
\( \implies a^2 = 8 \)
\( \implies a = \pm \sqrt{8} = \pm 2\sqrt{2} \)
When \( a = 2\sqrt{2} \), \( b = \frac{4}{2\sqrt{2}} = \sqrt{2} \).
When \( a = -2\sqrt{2} \), \( b = -\sqrt{2} \).
Thus, the square roots of \( 6 + 8i \) are \( \pm(2\sqrt{2} + \sqrt{2}i) \).
In simple words: To find the square root of a complex number, we assume it is equal to another complex number \( a + bi \). By squaring both sides and comparing the real and imaginary parts, we get simple quadratic equations to solve for \( a \) and \( b \).

🎯 Exam Tip: Always remember that since \( a \) is a real number, \( a^2 \) can never be negative. Discard the negative value of \( a^2 \) immediately to avoid calculation errors.

Question 1. Find the square root of the following complex numbers:
(i) \( \sqrt{-16 + 30i} \)
(ii) \( \sqrt{15 - 8i} \)
(iii) \( \sqrt{2 - 2\sqrt{3}i} \)
Answer:
(i)
\( \therefore a = \pm 3 \)
When \( a = 3 \), \( b = \frac{15}{3} = 5 \)
When \( a = -3 \), \( b = \frac{15}{-3} = -5 \)
\( \therefore \sqrt{-16 + 30i} = \pm(3 + 5i) \)

(ii) Let \( \sqrt{15 - 8i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 15 - 8i = a^2 + b^2i^2 + 2abi \)
\( \therefore 15 - 8i = (a^2 - b^2) + 2abi \) ...... [\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 15 \) and \( 2ab = -8 \)
\( \therefore a^2 - b^2 = 15 \) and \( b = \frac{-4}{a} \)
\( \therefore a^2 - \left(\frac{-4}{a}\right)^2 = 15 \)
\( \therefore a^2 - \frac{16}{a^2} = 15 \)
\( \dots \text{Multiplying by } a^2 \text{ on both sides} \)
\( \therefore a^4 - 16 = 15a^2 \)
\( \therefore a^4 - 15a^2 - 16 = 0 \)
\( \therefore (a^2 - 16)(a^2 + 1) = 0 \)
\( \dots a^2 = 16 \) or \( a^2 = -1 \)
But \( a \in \mathbb{R} \), \( a^2 \neq -1 \)
\( \therefore a^2 = 16 \)
\( \therefore a = \pm 4 \)
When \( a = 4 \), \( b = \frac{-4}{4} = -1 \)
When \( a = -4 \), \( b = \frac{-4}{-4} = 1 \)
\( \therefore \sqrt{15 - 8i} = \pm(4 - i) \)

(iii) Let \( \sqrt{2 - 2\sqrt{3}i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 2 - 2\sqrt{3}i = a^2 + b^2i^2 + 2abi \)
\( \therefore 2 - 2\sqrt{3}i = (a^2 - b^2) + 2abi \) ...... [\( \dots i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 2 \) and \( 2ab = -2\sqrt{3} \)
\( \therefore a^2 - b^2 = 2 \) and \( b = \frac{-\sqrt{3}}{a} \)
\( \therefore a^2 - \left(\frac{-\sqrt{3}}{a}\right)^2 = 2 \)
\( \dots a^2 - \frac{3}{a^2} = 2 \)
\( \therefore a^4 - 3 = 2a^2 \)
\( \therefore a^4 - 2a^2 - 3 = 0 \)
\( \therefore (a^2 - 3)(a^2 + 1) = 0 \)
\( \therefore a^2 = 3 \) or \( a^2 = -1 \)
But \( a \in \mathbb{R} \), \( a^2 \neq -1 \)
\( \therefore a^2 = 3 \)
\( \therefore a = \pm\sqrt{3} \)
When \( a = \sqrt{3} \), \( b = \frac{-\sqrt{3}}{\sqrt{3}} = -1 \)
When \( a = -\sqrt{3} \), \( b = \frac{-\sqrt{3}}{-\sqrt{3}} = 1 \)
\( \therefore \sqrt{2 - 2\sqrt{3}i} = \pm(\sqrt{3} - i) \)
In simple words: To find the square root of a complex number, we assume it equals \( a + bi \), square both sides, and solve for the real numbers \( a \) and \( b \). This gives us two values representing the positive and negative roots.

🎯 Exam Tip: Always remember that \( a \) and \( b \) must be real numbers. Therefore, you must discard any negative values for \( a^2 \) (such as \( a^2 = -1 \)) because the square of a real number is always non-negative.

Question 1. Find the square root of \( 2 + 2\sqrt{3}i \) (Continuation of solution)
Answer:
\( \therefore a^4 - 3 = 2a^2 \)
\( \therefore a^4 - 2a^2 - 3 = 0 \)
\( \dots (a^2 - 3)(a^2 + 1) = 0 \)
\( \therefore a^2 = 3 \) or \( a^2 = -1 \)
But \( a \in \mathbb{R} \), \( a^2 \neq -1 \)
\( \therefore a^2 = 3 \)
\( \therefore a = \pm\sqrt{3} \)
When \( a = \sqrt{3} \), \( b = \frac{\sqrt{3}}{\sqrt{3}} = 1 \)
When \( a = -\sqrt{3} \), \( b = \frac{\sqrt{3}}{-\sqrt{3}} = -1 \)
\( \therefore \sqrt{2 + 2\sqrt{3}i} = \pm(\sqrt{3} + i) \)
In simple words: We solve the quadratic-like equation for \( a^2 \), discard the imaginary solution since \( a \) must be a real number, and then find the corresponding values of \( b \) to get the final square roots.

🎯 Exam Tip: Remember that since \( a \) is a real number, \( a^2 \) cannot be negative. Always discard the negative value of \( a^2 \) with a proper reason to avoid losing marks.

Question 2. Find the square root of \( 18i \)
Answer:
Let \( \sqrt{18i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 18i = a^2 + b^2i^2 + 2abi \)
\( \therefore 0 + 18i = a^2 - b^2 + 2abi \) .....[\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 0 \) and \( 2ab = 18 \)
\( \therefore a^2 - b^2 = 0 \) and \( b = \frac{9}{a} \)
\( \therefore a^2 - \left(\frac{9}{a}\right)^2 = 0 \)
\( \therefore a^2 - \frac{81}{a^2} = 0 \)
\( \therefore a^4 - 81 = 0 \)
\( \therefore (a^2 - 9)(a^2 + 9) = 0 \)
\( \therefore a^2 = 9 \) or \( a^2 = -9 \)
But \( a \in \mathbb{R} \), \( a^2 \neq -9 \)
\( \therefore a^2 = 9 \)
\( \dots a = \pm3 \)
When \( a = 3 \), \( b = \frac{9}{3} = 3 \)
When \( a = -3 \), \( b = \frac{9}{-3} = -3 \)
\( \therefore \sqrt{18i} = \pm3(1 + i) \)
In simple words: To find the square root of \( 18i \), we set it equal to \( a + bi \), square both sides, and solve the resulting equations for the real numbers \( a \) and \( b \).

🎯 Exam Tip: When equating real and imaginary parts, don't forget to write the condition \( a, b \in \mathbb{R} \) as it justifies why \( a^2 \neq -9 \).

Question 3. Find the square root of \( 3 - 4i \)
Answer:
Let \( \sqrt{3 - 4i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 3 - 4i = a^2 + b^2i^2 + 2abi \)
\( \therefore 3 - 4i = a^2 - b^2 + 2abi \) ......[\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 3 \) and \( 2ab = -4 \)
\( \therefore a^2 - b^2 = 3 \) and \( b = \frac{-2}{a} \)
In simple words: We assume the square root is \( a + bi \), square both sides, and equate the real parts to \( 3 \) and the imaginary parts to \( -4 \) to set up our system of equations.

🎯 Exam Tip: Be careful with signs when equating the imaginary parts; here, \( 2ab = -4 \), which means \( b \) and \( a \) must have opposite signs.

Question 1(v). Find the square root of \( 3 - 4i \)
Answer:
\( \therefore a^2 - \left(-\frac{2}{a}\right)^2 = 3 \)
\( \therefore a^2 - \frac{4}{a^2} = 3 \)
\( \dots \)
\( \therefore a^4 - 4 = 3a^2 \)
\( \therefore a^4 - 3a^2 - 4 = 0 \)
\( \dots \)
\( \therefore (a^2 - 4)(a^2 + 1) = 0 \)
\( \therefore a^2 = 4 \) or \( a^2 = -1 \)
But, \( a \in \mathbb{R} \), \( a^2 \neq -1 \)
\( \therefore a^2 = 4 \)
\( \therefore a = \pm 2 \)
When \( a = 2 \), \( b = \frac{-2}{2} = -1 \)
When \( a = -2 \), \( b = \frac{-2}{-2} = 1 \)
\( \therefore \sqrt{3 - 4i} = \pm(2 - i) \)
In simple words: We solve the quadratic-like equation for \(a^2\), find the real values of \(a\), and then find the corresponding values of \(b\) to get the final square roots.

🎯 Exam Tip: Since \(a\) is a real number, its square \(a^2\) cannot be negative. Always discard the negative value of \(a^2\) and state this reason clearly to score full marks.

Question 1(vi). Find the square root of \( 6 + 8i \)
Answer:
Let \( \sqrt{6 + 8i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 6 + 8i = a^2 + b^2i^2 + 2abi \)
\( \therefore 6 + 8i = a^2 - b^2 + 2abi \) ......[\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 6 \) and \( 2ab = 8 \)
\( \therefore a^2 - b^2 = 6 \) and \( b = \frac{4}{a} \)
\( \therefore a^2 - \left(\frac{4}{a}\right)^2 = 6 \)
\( \therefore a^2 - \frac{16}{a^2} = 6 \)
\( \therefore a^4 - 16 = 6a^2 \)
\( \therefore a^4 - 6a^2 - 16 = 0 \)
\( \therefore (a^2 - 8)(a^2 + 2) = 0 \)
\( \therefore a^2 = 8 \) or \( a^2 = -2 \)
But \( a \in \mathbb{R} \), \( a^2 \neq -2 \)
\( \dots \)
\( \therefore a^2 = 8 \)
\( \therefore a = \pm 2\sqrt{2} \)
When \( a = 2\sqrt{2} \), \( b = \frac{4}{2\sqrt{2}} = \sqrt{2} \)
When \( a = -2\sqrt{2} \), \( b = \frac{4}{-2\sqrt{2}} = -\sqrt{2} \)
\( \therefore \sqrt{6 + 8i} = \pm(2\sqrt{2} + \sqrt{2}i) = \pm\sqrt{2}(2 + i) \)
In simple words: To find the square root of \(6 + 8i\), we assume it equals \(a + bi\), square both sides to set up equations for the real and imaginary parts, and solve for \(a\) and \(b\).

🎯 Exam Tip: When simplifying fractions like \(\frac{4}{2\sqrt{2}}\), rationalize the denominator to get \(\sqrt{2}\) quickly and avoid calculation errors.