GSEB Class 11 Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations G Exercise 5.1

Express each of the complex numbers given in questions 1 to 10 in the form a + ib:

Question 1. \( 5i \times (-\frac{3}{5}i) \)
Answer: We begin by multiplying the given complex numbers:
\( (5i) \left(-\frac{3}{5}i\right) \)
We can rearrange the terms as:
\( = \left(- 5 \times \frac{3}{5}\right) \times (i \times i) \)
\( = - 3i^2 \)
Since \( i^2 = -1 \), we substitute this value:
\( = (-3)(- 1) \)
\( = 3 \)
To write this in the standard form \( a + ib \), we express 3 as \( 3 + 0i \).
So, \( a = 3 \) and \( b = 0 \).
In simple words: When you multiply these two complex numbers, the result is 3. In the form \( a + ib \), this means a is 3 and b is 0.

Exam Tip: Remember that \( i^2 = -1 \) is a fundamental property for simplifying complex number expressions. Always convert the final answer to the \( a+ib \) format, even if \( b=0 \).

Question 2. \( i^9 + i^{19} \)
Answer: We need to simplify the powers of \( i \). We know that \( i^4 = 1 \).
\( i^9 = i^{4 \times 2 + 1} = (i^4)^2 \times i^1 = 1^2 \times i = i \)
\( i^{19} = i^{4 \times 4 + 3} = (i^4)^4 \times i^3 = 1^4 \times i^3 = i^3 \)
We also know that \( i^3 = -i \). So, \( i^{19} = -i \).
Now, substitute these simplified values back into the expression:
\( i^9 + i^{19} = i + (-i) \)
\( = i - i \)
\( = 0 \)
To write this in the standard form \( a + ib \), we express 0 as \( 0 + 0i \).
So, \( a = 0 \) and \( b = 0 \).
In simple words: First, simplify each part of the expression by dividing the exponent by 4 and seeing the remainder. Then, add the simplified terms. The final answer is 0.

Exam Tip: To simplify higher powers of \( i \), divide the exponent by 4 and use the remainder (0, 1, 2, or 3) to determine if it's \( i^0=1 \), \( i^1=i \), \( i^2=-1 \), or \( i^3=-i \).

Question 3. \( i^{-39} \)
Answer: We need to simplify the negative power of \( i \).
\( i^{-39} = \frac{1}{i^{39}} \)
First, let's simplify \( i^{39} \). We divide 39 by 4:
\( 39 = 4 \times 9 + 3 \)
So, \( i^{39} = i^3 \). We know that \( i^3 = -i \).
Substitute this back into the expression:
\( = \frac{1}{-i} \)
To get rid of \( i \) in the denominator, multiply both the numerator and the denominator by \( i \):
\( = \frac{1}{-i} \times \frac{i}{i} \)
\( = \frac{i}{-i^2} \)
Since \( i^2 = -1 \), we substitute this value:
\( = \frac{i}{-(-1)} \)
\( = \frac{i}{1} \)
\( = i \)
To write this in the standard form \( a + ib \), we express \( i \) as \( 0 + 1i \).
So, \( a = 0 \) and \( b = 1 \).
In simple words: To simplify negative powers of i, first turn it into a fraction with a positive power. Then, simplify the power of i. Finally, multiply the top and bottom by i to remove i from the denominator. The result is \( i \).

Exam Tip: When simplifying complex fractions, always rationalize the denominator by multiplying by the conjugate or by \( i \) if it's a simple \( i \) term. Ensure the final result is in \( a+ib \) form.

Question 4. \( 3(7 + i7) + i(7 + i7) \)
Answer: We need to expand and simplify the given expression:
\( 3(7 + i7) + i(7 + i7) \)
Distribute the 3 and the \( i \) into their respective parentheses:
\( = (3 \times 7 + 3 \times i7) + (i \times 7 + i \times i7) \)
\( = (21 + 21i) + (7i + 7i^2) \)
Since \( i^2 = -1 \), we substitute this value:
\( = (21 + 21i) + (7i + 7(-1)) \)
\( = (21 + 21i) + (7i - 7) \)
Now, group the real parts and the imaginary parts together:
\( = (21 - 7) + (21i + 7i) \)
\( = 14 + 28i \)
This is already in the standard form \( a + ib \).
So, \( a = 14 \) and \( b = 28 \).
In simple words: Expand both parts of the expression, remember that \( i^2 \) becomes -1, then group all the normal numbers together and all the numbers with 'i' together. The answer is \( 14 + 28i \).

Exam Tip: Be careful with signs, especially when \( i^2 \) changes a term from imaginary to real. Always combine real parts with real parts and imaginary parts with imaginary parts separately.

Question 5. \( (1 - i) - (- 1 + i6) \)
Answer: We need to simplify the given expression by performing the subtraction.
\( (1 - i) - (- 1 + i6) \)
Distribute the negative sign to the terms inside the second parenthesis:
\( = 1 - i + 1 - i6 \)
Now, group the real parts and the imaginary parts together:
\( = (1 + 1) + (-i - i6) \)
\( = 2 + (-1 - 6)i \)
\( = 2 - 7i \)
This is already in the standard form \( a + ib \).
So, \( a = 2 \) and \( b = -7 \).
In simple words: Remove the parentheses, remembering to change signs for the terms after the minus sign. Then, add the normal numbers together and the 'i' numbers together. The answer is \( 2 - 7i \).

Exam Tip: When subtracting complex numbers, always distribute the negative sign to every term in the second complex number before combining like terms.

Question 6. \( (\frac{1}{5} + i\frac{2}{5}) - (4 + i\frac{5}{2}) \)
Answer: We need to simplify the given expression by performing the subtraction.
\( \left(\frac{1}{5} + i\frac{2}{5}\right) - \left(4 + i\frac{5}{2}\right) \)
First, remove the parentheses and distribute the negative sign:
\( = \frac{1}{5} + \frac{2}{5}i - 4 - \frac{5}{2}i \)
Now, group the real parts and the imaginary parts together:
\( = \left(\frac{1}{5} - 4\right) + \left(\frac{2}{5}i - \frac{5}{2}i\right) \)
For the real part, find a common denominator (5):
\( \frac{1}{5} - \frac{4 \times 5}{5} = \frac{1}{5} - \frac{20}{5} = \frac{1 - 20}{5} = -\frac{19}{5} \)
For the imaginary part, find a common denominator (10):
\( \frac{2}{5}i - \frac{5}{2}i = \frac{2 \times 2}{10}i - \frac{5 \times 5}{10}i = \frac{4}{10}i - \frac{25}{10}i = \frac{4 - 25}{10}i = -\frac{21}{10}i \)
Combine these two results:
\( = -\frac{19}{5} - \frac{21}{10}i \)
This is in the standard form \( a + ib \).
So, \( a = -\frac{19}{5} \) and \( b = -\frac{21}{10} \).
In simple words: First, separate the real numbers and the imaginary numbers. Then, subtract the real parts by finding a common bottom number, and do the same for the imaginary parts. The answer is \( -\frac{19}{5} - \frac{21}{10}i \).

Exam Tip: Be careful when dealing with fractions. Always find a common denominator before adding or subtracting both the real and imaginary components.

Question 7. \( \left[\left(\frac{1}{3} + i\frac{7}{3}\right)\right] + \left(4 + i\frac{1}{3}\right) - \left(- \frac{4}{3} + i\right) \)
Answer: We need to simplify the given expression by performing the additions and subtraction.
\( \left(\frac{1}{3} + i\frac{7}{3}\right) + \left(4 + i\frac{1}{3}\right) - \left(- \frac{4}{3} + i\right) \)
First, remove the parentheses and distribute the negative sign:
\( = \frac{1}{3} + \frac{7}{3}i + 4 + \frac{1}{3}i + \frac{4}{3} - i \)
Now, group all the real parts and all the imaginary parts together:
\( = \left(\frac{1}{3} + 4 + \frac{4}{3}\right) + \left(\frac{7}{3}i + \frac{1}{3}i - i\right) \)
For the real part, find a common denominator (3):
\( \frac{1}{3} + \frac{4 \times 3}{3} + \frac{4}{3} = \frac{1}{3} + \frac{12}{3} + \frac{4}{3} = \frac{1 + 12 + 4}{3} = \frac{17}{3} \)
For the imaginary part, find a common denominator (3):
\( \frac{7}{3}i + \frac{1}{3}i - \frac{1 \times 3}{3}i = \frac{7}{3}i + \frac{1}{3}i - \frac{3}{3}i = \frac{7 + 1 - 3}{3}i = \frac{5}{3}i \)
Combine these two results:
\( = \frac{17}{3} + \frac{5}{3}i \)
This is in the standard form \( a + ib \).
So, \( a = \frac{17}{3} \) and \( b = \frac{5}{3} \).
In simple words: First, open all the brackets, changing signs where a minus is in front. Then, collect all the normal numbers to add them up, and collect all the 'i' numbers to add them up. Make sure to use common denominators for fractions. The result is \( \frac{17}{3} + \frac{5}{3}i \).

Exam Tip: When working with multiple complex numbers, it's often easiest to distribute all signs first, then collect all real terms and all imaginary terms before performing additions and subtractions.

Question 8. \( (1 - i)^4 \)
Answer: We need to expand and simplify the given expression.
\( (1 - i)^4 \)
We can write this as \( ((1 - i)^2)^2 \).
First, let's expand \( (1 - i)^2 \):
\( (1 - i)^2 = 1^2 - 2(1)(i) + i^2 \)
\( = 1 - 2i + (-1) \)
\( = 1 - 2i - 1 \)
\( = -2i \)
Now, substitute this back into the original expression:
\( (1 - i)^4 = (-2i)^2 \)
\( = (-2)^2 \times i^2 \)
\( = 4 \times (-1) \)
\( = -4 \)
To write this in the standard form \( a + ib \), we express -4 as \( -4 + 0i \).
So, \( a = -4 \) and \( b = 0 \).
In simple words: To find the fourth power, first find the square of the expression, then square that result. Remember that \( i^2 \) is -1. The final answer is -4.

Exam Tip: When dealing with higher powers of binomials like \( (a+b)^n \), try to break them down into smaller, easier-to-manage powers, like squaring a squared term, as \( ((a+b)^2)^2 \).

Question 9. \( \left(\frac{1}{3} + 3i\right)^3 \)
Answer: We need to expand the given expression using the binomial formula \( (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \).
Here, \( x = \frac{1}{3} \) and \( y = 3i \).
\( \left(\frac{1}{3} + 3i\right)^3 = \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2 (3i) + 3\left(\frac{1}{3}\right)(3i)^2 + (3i)^3 \)
Calculate each term:
\( \left(\frac{1}{3}\right)^3 = \frac{1}{27} \)
\( 3\left(\frac{1}{3}\right)^2 (3i) = 3\left(\frac{1}{9}\right)(3i) = \frac{9}{9}i = i \)
\( 3\left(\frac{1}{3}\right)(3i)^2 = 3\left(\frac{1}{3}\right)(9i^2) = 9i^2 \)
Since \( i^2 = -1 \), \( 9i^2 = 9(-1) = -9 \).
\( (3i)^3 = 3^3 i^3 = 27i^3 \)
Since \( i^3 = -i \), \( 27i^3 = 27(-i) = -27i \).
Now, combine all the terms:
\( = \frac{1}{27} + i - 9 - 27i \)
Group the real parts and the imaginary parts:
\( = \left(\frac{1}{27} - 9\right) + (i - 27i) \)
For the real part:
\( \frac{1}{27} - \frac{9 \times 27}{27} = \frac{1 - 243}{27} = -\frac{242}{27} \)
For the imaginary part:
\( 1i - 27i = -26i \)
Combine these two results:
\( = -\frac{242}{27} - 26i \)
This is in the standard form \( a + ib \).
So, \( a = -\frac{242}{27} \) and \( b = -26 \).
In simple words: Use the cube formula to expand the expression. Simplify each part, remembering \( i^2 = -1 \) and \( i^3 = -i \). Then, gather all the normal numbers and all the 'i' numbers to get the final answer.

Exam Tip: When expanding binomials with imaginary numbers, always simplify powers of \( i \) (especially \( i^2 \) and \( i^3 \)) immediately to convert terms from imaginary to real or vice versa, making it easier to combine like terms.

Question 10. \( \left(-2-\frac{1}{3} i\right)^3 \)
Answer: We need to expand the given expression using the binomial formula \( (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 \).
Here, \( x = -2 \) and \( y = -\frac{1}{3}i \).
\( \left(-2-\frac{1}{3}i\right)^3 = (-2)^3 + 3(-2)^2\left(-\frac{1}{3}i\right) + 3(-2)\left(-\frac{1}{3}i\right)^2 + \left(-\frac{1}{3}i\right)^3 \)
Calculate each term:
\( (-2)^3 = -8 \)
\( 3(-2)^2\left(-\frac{1}{3}i\right) = 3(4)\left(-\frac{1}{3}i\right) = 12\left(-\frac{1}{3}i\right) = -4i \)
\( 3(-2)\left(-\frac{1}{3}i\right)^2 = -6\left(\frac{1}{9}i^2\right) = -\frac{6}{9}i^2 = -\frac{2}{3}i^2 \)
Since \( i^2 = -1 \), \( -\frac{2}{3}i^2 = -\frac{2}{3}(-1) = \frac{2}{3} \).
\( \left(-\frac{1}{3}i\right)^3 = \left(-\frac{1}{3}\right)^3 i^3 = -\frac{1}{27}i^3 \)
Since \( i^3 = -i \), \( -\frac{1}{27}i^3 = -\frac{1}{27}(-i) = \frac{1}{27}i \).
Now, combine all the terms:
\( = -8 - 4i + \frac{2}{3} + \frac{1}{27}i \)
Group the real parts and the imaginary parts:
\( = \left(-8 + \frac{2}{3}\right) + \left(-4i + \frac{1}{27}i\right) \)
For the real part:
\( -8 + \frac{2}{3} = -\frac{8 \times 3}{3} + \frac{2}{3} = -\frac{24}{3} + \frac{2}{3} = \frac{-24 + 2}{3} = -\frac{22}{3} \)
For the imaginary part:
\( -4i + \frac{1}{27}i = -\frac{4 \times 27}{27}i + \frac{1}{27}i = -\frac{108}{27}i + \frac{1}{27}i = \frac{-108 + 1}{27}i = -\frac{107}{27}i \)
Combine these two results:
\( = -\frac{22}{3} - \frac{107}{27}i \)
This is in the standard form \( a + ib \).
So, \( a = -\frac{22}{3} \) and \( b = -\frac{107}{27} \).
In simple words: Expand the expression using the cube formula, being careful with negative signs and fractions. Simplify powers of 'i' as you go. Group the normal numbers and the 'i' numbers separately to get the final complex number.

Exam Tip: When dealing with negative signs within binomial expansion, be meticulous with each term's sign. Squaring a negative gives a positive, while cubing a negative keeps it negative. This also applies to the imaginary unit \( i \).

Find the multiplicative inverse of each of the complex numbers given in questions 11 to 13:

Question 11. \( 4-3i \)
Answer: To find the multiplicative inverse of a complex number \( z = a + bi \), we calculate \( \frac{1}{z} = \frac{1}{a + bi} \). We then multiply the numerator and denominator by the conjugate of the denominator, which is \( a - bi \).
For \( z = 4 - 3i \), the multiplicative inverse is \( \frac{1}{4 - 3i} \).
The conjugate of \( 4 - 3i \) is \( 4 + 3i \).
\( = \frac{1}{4 - 3i} \times \frac{4 + 3i}{4 + 3i} \)
In the denominator, we use the formula \( (x - y)(x + y) = x^2 - y^2 \):
\( = \frac{4 + 3i}{4^2 - (3i)^2} \)
\( = \frac{4 + 3i}{16 - 9i^2} \)
Since \( i^2 = -1 \), we substitute this value:
\( = \frac{4 + 3i}{16 - 9(-1)} \)
\( = \frac{4 + 3i}{16 + 9} \)
\( = \frac{4 + 3i}{25} \)
Now, separate the real and imaginary parts to express in the form \( a + ib \):
\( = \frac{4}{25} + \frac{3}{25}i \)
So, the multiplicative inverse is \( \frac{4}{25} + \frac{3}{25}i \).
In simple words: To find the inverse, put 1 over the complex number. Then, multiply the top and bottom by the complex conjugate of the bottom part. Simplify using \( i^2 = -1 \), and write the answer as a normal number plus an 'i' number.

Exam Tip: The key to finding a multiplicative inverse is to multiply by the complex conjugate in the denominator to eliminate the imaginary part from the bottom of the fraction, thereby rationalizing it.

Question 12. \( \sqrt{5} + 3i \)
Answer: To find the multiplicative inverse of \( z = \sqrt{5} + 3i \), we calculate \( \frac{1}{z} = \frac{1}{\sqrt{5} + 3i} \). We then multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{5} - 3i \).
For \( z = \sqrt{5} + 3i \), the multiplicative inverse is \( \frac{1}{\sqrt{5} + 3i} \).
The conjugate of \( \sqrt{5} + 3i \) is \( \sqrt{5} - 3i \).
\( = \frac{1}{\sqrt{5} + 3i} \times \frac{\sqrt{5} - 3i}{\sqrt{5} - 3i} \)
In the denominator, we use the formula \( (x + y)(x - y) = x^2 - y^2 \):
\( = \frac{\sqrt{5} - 3i}{(\sqrt{5})^2 - (3i)^2} \)
\( = \frac{\sqrt{5} - 3i}{5 - 9i^2} \)
Since \( i^2 = -1 \), we substitute this value:
\( = \frac{\sqrt{5} - 3i}{5 - 9(-1)} \)
\( = \frac{\sqrt{5} - 3i}{5 + 9} \)
\( = \frac{\sqrt{5} - 3i}{14} \)
Now, separate the real and imaginary parts to express in the form \( a + ib \):
\( = \frac{\sqrt{5}}{14} - \frac{3}{14}i \)
So, the multiplicative inverse is \( \frac{\sqrt{5}}{14} - \frac{3}{14}i \).
In simple words: To get the inverse, make it a fraction with 1 on top. Then, multiply the top and bottom by the complex conjugate of the bottom. Simplify the expression, remembering \( i^2 = -1 \), and present it in the \( a + ib \) format.

Exam Tip: Be mindful of square roots in the real part of the complex number. They are treated just like any other real number when finding the conjugate and squaring the denominator.

Question 13. \( -i \)
Answer: To find the multiplicative inverse of \( z = -i \), we calculate \( \frac{1}{z} = \frac{1}{-i} \).
To simplify this expression and remove \( i \) from the denominator, multiply both the numerator and the denominator by \( i \):
\( = \frac{1}{-i} \times \frac{i}{i} \)
\( = \frac{i}{-i^2} \)
Since \( i^2 = -1 \), we substitute this value:
\( = \frac{i}{-(-1)} \)
\( = \frac{i}{1} \)
\( = i \)
To write this in the standard form \( a + ib \), we express \( i \) as \( 0 + 1i \).
So, the multiplicative inverse is \( 0 + i \).
In simple words: To find the inverse of \( -i \), put 1 over \( -i \). Then multiply the top and bottom by \( i \). Simplify, remembering \( i^2 = -1 \), and the answer is \( i \).

Exam Tip: When the denominator is simply \( i \) or \( -i \), multiplying by \( i/i \) is sufficient to rationalize the denominator, as \( i^2 = -1 \).

Question 14. Express the following expression in the form a + ib: \( \frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i \sqrt{2})} \)
Answer: We need to simplify the given complex expression and write it in the form \( a + ib \).
The expression is: \( \frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i \sqrt{2})} \)
First, let's simplify the numerator using the difference of squares formula \( (x+y)(x-y) = x^2-y^2 \):
Numerator: \( (3+i \sqrt{5})(3-i \sqrt{5}) = 3^2 - (i \sqrt{5})^2 \)
\( = 9 - i^2(\sqrt{5})^2 \)
\( = 9 - (-1)(5) \)
\( = 9 + 5 = 14 \)
Next, let's simplify the denominator:
Denominator: \( (\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i \sqrt{2}) \)
Distribute the negative sign:
\( = \sqrt{3}+\sqrt{2}i - \sqrt{3} + i \sqrt{2} \)
Group the real and imaginary parts:
\( = (\sqrt{3} - \sqrt{3}) + (\sqrt{2}i + \sqrt{2}i) \)
\( = 0 + (1\sqrt{2} + 1\sqrt{2})i \)
\( = 2\sqrt{2}i \)
Now, substitute the simplified numerator and denominator back into the original expression:
\( = \frac{14}{2\sqrt{2}i} \)
Simplify the fraction:
\( = \frac{7}{\sqrt{2}i} \)
To eliminate \( i \) from the denominator, multiply the numerator and denominator by \( i \):
\( = \frac{7}{\sqrt{2}i} \times \frac{i}{i} \)
\( = \frac{7i}{\sqrt{2}i^2} \)
Since \( i^2 = -1 \), substitute this value:
\( = \frac{7i}{\sqrt{2}(-1)} \)
\( = \frac{7i}{-\sqrt{2}} \)
\( = -\frac{7}{\sqrt{2}}i \)
To rationalize the denominator further, multiply the numerator and denominator by \( \sqrt{2} \):
\( = -\frac{7}{\sqrt{2}}i \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( = -\frac{7\sqrt{2}}{2}i \)
To write this in the standard form \( a + ib \), we express it as \( 0 - \frac{7\sqrt{2}}{2}i \).
So, \( a = 0 \) and \( b = -\frac{7\sqrt{2}}{2} \).
In simple words: First, simplify the top part using the difference of squares formula, and simplify the bottom part by combining like terms. Then, divide the top by the bottom. To get 'i' out of the bottom, multiply the top and bottom by 'i', and then by \( \sqrt{2} \) to rationalize. The final answer is \( 0 - \frac{7\sqrt{2}}{2}i \).

Exam Tip: Always simplify the numerator and denominator separately before combining them. Remember to rationalize denominators with imaginary numbers by multiplying by the conjugate or by \( i \).