CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations Assignment Set 02

Question 1. Evaluate, \(\sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625}\)
Answer: Given expression: \(\sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625}\)
Using the property \(\sqrt{-1} = i\), we get:
\(= \sqrt{16}i + 3\sqrt{25}i + \sqrt{36}i - \sqrt{625}i\)
\(= 4i + 3(5)i + 6i - 25i\)
\(= 4i + 15i + 6i - 25i\)
\(= 25i - 25i = 0\)
In simple words: Convert each negative square root into terms of \(i\) using \(\sqrt{-1} = i\) and simplify like terms.

Exam Tip: Always remember that \(\sqrt{-a} = i\sqrt{a}\) for positive real number \(a\), and be careful with signs.

Question 2. Evaluate, \(i^{29} + \frac{1}{i^{29}}\)
Answer: Given expression: \(i^{29} + \frac{1}{i^{29}}\)
Dividing 29 by 4, we get a quotient of 7 and a remainder of 1, so \(29 = 4(7) + 1\).
\(i^{29} = i^{4(7)+1} = (i^4)^7 \cdot i^1 = (1)^7 \cdot i = i\)
Similarly, \(\frac{1}{i^{29}} = \frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i\)
Sum = \(i + (-i) = 0\)
In simple words: Use powers of \(i\) where \(i^4 = 1\) to simplify large exponents by looking at the remainder upon division by 4.

Exam Tip: Recall that \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\).

Question 3. Find values of x and y if, \((3x - 7) + 2iy = -5y + (5 + x)i\)
Answer: Equating the real and imaginary parts of the given complex equation:
Real part: \(3x - 7 = -5y \implies 3x + 5y = 7\) --- (1)
Imaginary part: \(2y = 5 + x \implies x - 2y = -5\) --- (2)
From equation (2), we get \(x = 2y - 5\).
Substituting \(x\) into equation (1):
\(3(2y - 5) + 5y = 7\)
\(6y - 15 + 5y = 7\)
\(11y = 22 \implies y = 2\)
Now, substituting \(y = 2\) into equation (2):
\(x = 2(2) - 5 = 4 - 5 = -1\)
Thus, \(x = -1\) and \(y = 2\).
In simple words: Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Exam Tip: Always separate real and imaginary parts clearly before setting up simultaneous linear equations.

Question 4. Express \(\frac{i}{1 + i}\) in the form \(a + ib\)
Answer: Given expression: \(\frac{i}{1 + i}\)
Multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 - i\):
\(= \frac{i(1 - i)}{(1 + i)(1 - i)}\)
\(= \frac{i - i^2}{1^2 - i^2}\)
Since \(i^2 = -1\):
\(= \frac{i - (-1)}{1 - (-1)}\)
\(= \frac{1 + i}{2} = \frac{1}{2} + \frac{1}{2}i\)
In simple words: Rationalize the denominator by multiplying top and bottom by the conjugate of the denominator.

Exam Tip: The conjugate of \(a + ib\) is \(a - ib\), and \((a+ib)(a-ib) = a^2 + b^2\).

Question 5. If \(z = \frac{1}{3 + 4i}\), find the conjugate of \(z\)
Answer: First, simplify \(z\) into the standard form \(a + ib\):
\(z = \frac{1}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i}\)
\(z = \frac{3 - 4i}{3^2 - (4i)^2} = \frac{3 - 4i}{9 - 16(-1)} = \frac{3 - 4i}{9 + 16} = \frac{3 - 4i}{25} = \frac{3}{25} - \frac{4}{25}i\)
The conjugate of \(z\), denoted as \(\bar{z}\), is obtained by changing the sign of the imaginary part:
\(\bar{z} = \frac{3}{25} + \frac{4}{25}i\)
In simple words: First simplify the complex number into standard form, then flip the sign of the imaginary part to find the conjugate.

Exam Tip: Alternatively, use the property of conjugates: \(\left(\frac{z_1}{z_2}\right) = \frac{\bar{z}_1}{\bar{z}_2}\).

Question 6. Find the modulus of \(z = 3 - 2i\)
Answer: Given \(z = 3 - 2i\), here real part \(a = 3\) and imaginary part \(b = -2\).
The modulus is given by \(|z| = \sqrt{a^2 + b^2}\):
\(|z| = \sqrt{3^2 + (-2)^2}\)
\(|z| = \sqrt{9 + 4} = \sqrt{13}\)
In simple words: Modulus is the distance of the complex number from the origin, calculated as \(\sqrt{\text{real}^2 + \text{imaginary}^2}\).

Exam Tip: Modulus is always a non-real negative-free positive real number.

Question 7. If \(z\) is a purely imaginary number and lies on the positive direction of y-axis then what is the argument of \(z\)?
Answer: A purely imaginary number lying on the positive direction of the y-axis is of the form \(z = iy = |y|i\) where \(y > 0\).
In the complex plane, this corresponds to the point \((0, y)\) on the positive y-axis.
The angle made with the positive direction of the x-axis is \(\frac{\pi}{2}\).
Therefore, the argument of \(z\) is \(\frac{\pi}{2}\).
In simple words: The positive y-axis points straight up, making a \(90^\circ\) or \(\frac{\pi}{2}\) angle with the positive x-axis.

Exam Tip: Positive x-axis has argument \(0\), positive y-axis has \(\frac{\pi}{2}\), negative x-axis has \(\pi\), and negative y-axis has \(-\frac{\pi}{2}\).

Question 8. Find the multiplicative inverse of \(5 + 3i\)
Answer: Let \(z = 5 + 3i\). The multiplicative inverse is \(z^{-1} = \frac{1}{z} = \frac{\bar{z}}{|z|^2}\).
Here, \(\bar{z} = 5 - 3i\)
\(|z|^2 = 5^2 + 3^2 = 25 + 9 = 34\)
Therefore:
\(z^{-1} = \frac{5 - 3i}{34} = \frac{5}{34} - \frac{3}{34}i\)
In simple words: Multiplicative inverse is simply the reciprocal of the complex number, simplified by multiplying numerator and denominator by its conjugate.

Question 9. If \(|z| = 4\) and argument of \(z = \frac{5\pi}{6}\) then write \(z\) in the form \(x + iy\); \(x, y \in \mathbb{R}\)
Answer: The polar form of a complex number is given by \(z = r(\cos\theta + i\sin\theta)\), where \(r = |z|\) and \(\theta = \text{arg}(z)\).
Here, \(r = 4\) and \(\theta = \frac{5\pi}{6}\).
\(z = 4\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)\)
We know that \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\) and \(\sin\frac{5\pi}{6} = \frac{1}{2}\).
\(z = 4\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right)\)
\(z = -2\sqrt{3} + 2i\)
In simple words: Convert from polar form to Cartesian form by evaluating the sine and cosine values of the given angle.

Exam Tip: \(\frac{5\pi}{6}\) lies in the second quadrant, where cosine is negative and sine is positive.

Question 10. If \(z = 1 - i\), find \(\text{Im}\left(\frac{1}{z\bar{z}}\right)\)
Answer: Given \(z = 1 - i\).
We know that \(z\bar{z} = |z|^2\).
For \(z = 1 - i\), \(|z|^2 = 1^2 + (-1)^2 = 1 + 1 = 2\).
Thus, \(\frac{1}{z\bar{z}} = \frac{1}{2}\), which is a purely real number.
Therefore, its imaginary part is \(0\).
\(\text{Im}\left(\frac{1}{z\bar{z}}\right) = 0\)
In simple words: The product of a complex number and its conjugate is always a real number, so its imaginary part is zero.

Exam Tip: Remember the fundamental property: \(z\bar{z} = |z|^2\).

Question 11. Simplify \((-i)(3i)\left(\frac{-1}{6}i\right)^3\)
Answer: Given expression: \((-i)(3i)\left(-\frac{1}{6}i\right)^3\)
First, simplify the first two terms: \((-i)(3i) = -3i^2 = -3(-1) = 3\).
Now, expand the cubed term:
\(\left(-\frac{1}{6}i\right)^3 = \left(-\frac{1}{6}\right)^3 i^3 = -\frac{1}{216}(-i) = \frac{i}{216}\)
Multiply them together:
\(3 \cdot \frac{i}{216} = \frac{3i}{216} = \frac{i}{72}\)
In simple words: Simplify the terms step-by-step using \(i^2 = -1\) and \(i^3 = -i\).

Exam Tip: Be careful with cube signs and powers of fractions.

Question 12. Find the solution of the equation \(x^2 + 5 = 0\) in complex numbers.
Answer: Given equation: \(x^2 + 5 = 0\)
\(x^2 = -5\)
\(x = \pm\sqrt{-5}\)
\(x = \pm\sqrt{5}i\)
In simple words: Move the constant to the right side and take the square root, replacing the negative sign with \(i\).

Exam Tip: Quadratic equations with negative discriminants always yield conjugate pairs of complex roots.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 13. For Complex numbers \(z_1 = -1 + i\), \(z_2 = 3 - 2i\) show that, \(\text{Im}(z_1z_2) = \text{Re}(z_1)\text{Im}(z_2) + \text{Im}(z_1)\text{Re}(z_2)\)
Answer: Given \(z_1 = -1 + i\) and \(z_2 = 3 - 2i\).
First, find the product \(z_1z_2\):
\(z_1z_2 = (-1 + i)(3 - 2i)\)
\(z_1z_2 = -3 + 2i + 3i - 2i^2\)
Since \(i^2 = -1\):
\(z_1z_2 = -3 + 5i - 2(-1) = -3 + 5i + 2 = -1 + 5i\)
Thus, the imaginary part of the product is:
\(\text{Im}(z_1z_2) = 5\) --- (Left Hand Side)
Now, let us find the individual real and imaginary parts:
\(\text{Re}(z_1) = -1, \quad \text{Im}(z_1) = 1\)
\(\text{Re}(z_2) = 3, \quad \text{Im}(z_2) = -2\)
Calculate the Right Hand Side:
\(\text{Re}(z_1)\text{Im}(z_2) + \text{Im}(z_1)\text{Re}(z_2) = (-1)(-2) + (1)(3)\)
\(= 2 + 3 = 5\) --- (Right Hand Side)
Since L.H.S. = R.H.S., hence proved.
In simple words: Multiply the two complex numbers directly to find the imaginary part of the product, and verify it using the formula involving individual real and imaginary parts.

Exam Tip: This is a standard property of complex numbers: \(\text{Im}(z_1z_2) = x_1y_2 + y_1x_2\).

Question 14. Convert the complex number \(-3\sqrt{2} + 3\sqrt{2}i\) in polar form
Answer: Let \(z = -3\sqrt{2} + 3\sqrt{2}i\). Here \(a = -3\sqrt{2}\) and \(b = 3\sqrt{2}\).
Step 1: Find the modulus \(r\):
\(r = |z| = \sqrt{(-3\sqrt{2})^2 + (3\sqrt{2})^2} = \sqrt{18 + 18} = \sqrt{36} = 6\)
Step 2: Find the argument \(\theta\):
\(\cos\theta = \frac{a}{r} = \frac{-3\sqrt{2}}{6} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}\)
\(\sin\theta = \frac{b}{r} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\)
Since \(\cos\theta\) is negative and \(\sin\theta\) is positive, \(z\) lies in the second quadrant.
The acute angle \(\alpha\) is given by \(\tan\alpha = \left|\frac{b}{a}\right| = \left|\frac{3\sqrt{2}}{-3\sqrt{2}}\right| = 1 \implies \alpha = \frac{\pi}{4}\).
Therefore, the principal argument is \(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Step 3: Write in polar form \(z = r(\cos\theta + i\sin\theta\)):
\(z = 6\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)
In simple words: Find the modulus \(r\) and argument \(\theta\) by identifying the correct quadrant, then substitute them into the polar form expression.

Exam Tip: Always check the quadrant based on the signs of \(a\) and \(b\) before finalizing the argument \(\theta\).

Question 15. If \(x + iy = \sqrt{\frac{1 + i}{1 - i}}\), prove that \(x^2 + y^2 = 1\)
Answer: Given \(x + iy = \sqrt{\frac{1 + i}{1 - i}}\)
Take the modulus on both sides:
\(|x + iy| = \left|\sqrt{\frac{1 + i}{1 - i}}\right|\)
\(\sqrt{x^2 + y^2} = \frac{|1 + i|}{|1 - i|}\)
Calculate the moduli of numerator and denominator:
\(|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}\)
\(|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\)
Therefore:
\(\sqrt{x^2 + y^2} = \frac{\sqrt{2}}{\sqrt{2}} = 1\)
Squaring both sides:
\(x^2 + y^2 = 1\)
In simple words: Taking the modulus of both sides makes this problem very short and elegant without needing to rationalize inside the square root first.

Exam Tip: Use the property \(|z_1/z_2| = |z_1| / |z_2|\) to simplify modulus equations quickly.

Question 16. Find real value of \(\theta\) such that, \(\frac{1 + i\cos\theta}{1 - 2i\cos\theta}\) is a real number
Answer: Let \(z = \frac{1 + i\cos\theta}{1 - 2i\cos\theta}\).
To make \(z\) a real number, its imaginary part must be equal to \(0\).
First, rationalize the denominator:
\(z = \frac{(1 + i\cos\theta)(1 + 2i\cos\theta)}{(1 - 2i\cos\theta)(1 + 2i\cos\theta)}\)
\(z = \frac{1 + 2i\cos\theta + i\cos\theta + 2i^2\cos^2\theta}{1^2 - (2i\cos\theta)^2}\)
Since \(i^2 = -1\):
\(z = \frac{(1 - 2\cos^2\theta) + i(3\cos\theta)}{1 + 4\cos^2\theta}\)
\(z = \frac{1 - 2\cos^2\theta}{1 + 4\cos^2\theta} + i\frac{3\cos\theta}{1 + 4\cos^2\theta}\)
For \(z\) to be a real number, the imaginary part must be zero:
\(\frac{3\cos\theta}{1 + 4\cos^2\theta} = 0\)
Since the denominator is never zero, we have:
\(3\cos\theta = 0 \implies \cos\theta = 0\)
The general solution for \(\cos\theta = 0\) is:
\(\theta = (2n + 1)\frac{\pi}{2}, \quad n \in \mathbb{Z}\)
In simple words: Simplify the expression into standard form and set its imaginary part equal to zero to solve for \(\theta\).

Exam Tip: A complex number is purely real if and only if its imaginary part is \(0\).

Question 17. If \(\left|\frac{z - 5i}{z + 5i}\right| = 1\), show that \(z\) is a real number.
Answer: Given \(\left|\frac{z - 5i}{z + 5i}\right| = 1\)
Using the modulus property \(\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}\):
\(\frac{|z - 5i|}{|z + 5i|} = 1 \implies |z - 5i| = |z + 5i|\)
Let \(z = x + iy\). Then:
\(|x + iy - 5i| = |x + iy + 5i|\)
\(|x + i(y - 5)| = |x + i(y + 5)|\)
Squaring both sides:
\(x^2 + (y - 5)^2 = x^2 + (y + 5)^2\)
Subtracting \(x^2\) from both sides:
\((y - 5)^2 = (y + 5)^2\)
\(y^2 - 10y + 25 = y^2 + 10y + 25\)
\(-10y = 10y \implies 20y = 0 \implies y = 0\)
Since the imaginary part \(y = 0$, \(z = x + 0i = x\), which is a real number.
In simple words: Substitute \(z = x + iy\) into the modulus equation and equate distances to show that the imaginary component \(y\) must be zero.

Exam Tip: The locus of points equidistant from two fixed points \((0, 5)\) and \((0, -5)\) is the perpendicular bisector (the x-axis, where \(y = 0\)).

Question 18. If \((x + iy)^{\frac{1}{3}} = a + ib\), prove that, \(\frac{x}{a} + \frac{y}{b} = 4(a^2 - b^2)\)
Answer: Given \((x + iy)^{\frac{1}{3}} = a + ib\)
Cubing both sides:
\(x + iy = (a + ib)^3\)
Using the expansion \((a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3\):
\(x + iy = a^3 + 3a^2bi - 3ab^2 - b^3i\)
\(x + iy = (a^3 - 3ab^2) + i(3a^2b - b^3)\)
Equating real and imaginary parts:
\(x = a^3 - 3ab^2\)
\(y = 3a^2b - b^3\)
Now, let us find \(\frac{x}{a}\) and \(\frac{y}{b}\):
\(\frac{x}{a} = \frac{a(a^2 - 3b^2)}{a} = a^2 - 3b^2\) (since \(a \neq 0\))
\(\frac{y}{b} = \frac{b(3a^2 - b^2)}{b} = 3a^2 - b^2\) (since \(b \neq 0\))
Adding \(\frac{x}{a}\) and \(\frac{y}{b}\):
\(\frac{x}{a} + \frac{y}{b} = (a^2 - 3b^2) + (3a^2 - b^2)\)
\(= 4a^2 - 4b^2 = 4(a^2 - b^2)\)
Hence proved.
In simple words: Cube both sides, separate real and imaginary parts for \(x\) and \(y\), divide by \(a\) and \(b\) respectively, and add them up.

Exam Tip: Remember algebraic identity \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\).

Question 19. For complex numbers \(z_1 = 6 + 3i\), \(z_2 = 3 - i\) find \(\frac{z_1}{z_2}\)
Answer: Given \(z_1 = 6 + 3i\) and \(z_2 = 3 - i\).
\(\frac{z_1}{z_2} = \frac{6 + 3i}{3 - i}\)
Multiply numerator and denominator by the conjugate of \(z_2\), which is \(3 + i\):
\(= \frac{(6 + 3i)(3 + i)}{(3 - i)(3 + i)}\)
\(= \frac{18 + 6i + 9i + 3i^2}{3^2 - i^2}\)
Since \(i^2 = -1\):
\(= \frac{18 + 15i + 3(-1)}{9 - (-1)}\)
\(= \frac{15 + 15i}{10} = \frac{15}{10} + \frac{15}{10}i = \frac{3}{2} + \frac{3}{2}i\)
Alternatively expressed as: \(\frac{3}{2}(1 + i)\).
In simple words: Divide two complex numbers by rationalizing the denominator using the conjugate of the divisor.

Exam Tip: Always reduce the final complex fraction to its simplest lowest terms.

Question 20. If \(\left(\frac{2 + 2i}{2 - 2i}\right)^n = 1\), find the least positive integral value of \(n\).
Answer: Simplify the base term inside the parentheses:
\(\frac{2 + 2i}{2 - 2i} = \frac{2(1 + i)}{2(1 - i)} = \frac{1 + i}{1 - i}\)
Now, rationalize \(\frac{1 + i}{1 - i}\):
\(\frac{1 + i}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{(1 + i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i\)
Therefore, the given equation becomes:
\(i^n = 1\)
We know that the lowest positive integer \(n\) for which \(i^n = 1\) is \(4\) (since \(i^4 = 1\)).
Thus, the least positive integral value of \(n\) is \(4\).
In simple words: Simplify the fraction inside to simply \(i\), then find the smallest positive power of \(i\) that equals \(1\).

Exam Tip: Keep in mind that \(i^4 = 1\) is the fundamental cycle period for powers of iota.

Question 21. Find the modulus and argument of \(z = 2 - 2i\)
Answer: Given \(z = 2 - 2i\). Here \(a = 2\) and \(b = -2\).
Modulus \(r = |z| = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\).
Argument \(\theta\):
\(\cos\theta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}\)
\(\sin\theta = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}}\)
Since \(\cos\theta > 0\) and \(\sin\theta < 0\), \(z\) lies in the fourth quadrant.
The acute angle is \(\alpha = \frac{\pi}{4}\).
Therefore, the principal argument is \(\theta = -\alpha = -\frac{\pi}{4}\).
Modulus = \(2\sqrt{2}\), Argument = \(-\frac{\pi}{4}\).
In simple words: Modulus is calculated using Pythagoras theorem, and argument is found by determining the quadrant and acute angle.

Exam Tip: In the fourth quadrant, the principal argument is negative (\(-\alpha\)).

Question 22. Solve the equation, \(\sqrt{3}x^2 - \sqrt{2}x + 3\sqrt{3} = 0\)
Answer: Given quadratic equation: \(\sqrt{3}x^2 - \sqrt{2}x + 3\sqrt{3} = 0\)
Here, \(A = \sqrt{3}\), \(B = -\sqrt{2}\), and \(C = 3\sqrt{3}\).
Using the quadratic formula \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\):
Discriminant \(D = B^2 - 4AC = (-\sqrt{2})^2 - 4(\sqrt{3})(3\sqrt{3})\)
\(D = 2 - 4(3)(3) = 2 - 36 = -34\)
Since \(D < 0\), the roots are complex:
\(x = \frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2(\sqrt{3})}\)
\(x = \frac{\sqrt{2} \pm \sqrt{34}i}{2\sqrt{3}}\)
In simple words: Apply the standard quadratic formula and use \(i\) for the negative square root of the discriminant.

Exam Tip: Double-check coefficients carefully when radicals are involved.

LONG ANSWER TYPE QUESTIONS (6 MARKS)

Question 23. If \(z_1, z_2\) are complex numbers such that, \(\left|\frac{z_1 - 3z_2}{3 - z_1\bar{z}_2}\right| = 1\) and \(|z_2| \neq 1\) then find \(|z_1|\)
Answer: Given \(\left|\frac{z_1 - 3z_2}{3 - z_1\bar{z}_2}\right| = 1\)
This implies: \(|z_1 - 3z_2| = |3 - z_1\bar{z}_2|\)
Squaring both sides:
\(|z_1 - 3z_2|^2 = |3 - z_1\bar{z}_2|^2\)
Using the property \(|z|^2 = z\bar{z}\):
\((z_1 - 3z_2)(\bar{z}_1 - 3\bar{z}_2) = (3 - z_1\bar{z}_2)(3 - \bar{z}_1z_2)\)
Expanding both sides:
\(z_1\bar{z}_1 - 3z_1\bar{z}_2 - 3\bar{z}_1z_2 + 9z_2\bar{z}_2 = 9 - 3\bar{z}_1z_2 - 3z_1\bar{z}_2 + z_1\bar{z}_1 z_2\bar{z}_2\)
Cancel out the common terms \(-3z_1\bar{z}_2 - 3\bar{z}_1z_2\) from both sides:
\(|z_1|^2 + 9|z_2|^2 = 9 + |z_1|^2|z_2|^2\)
Rearranging terms:
\(|z_1|^2 - 9 - |z_1|^2|z_2|^2 + 9|z_2|^2 = 0\)
\(|z_1|^2(1 - |z_2|^2) - 9(1 - |z_2|^2) = 0\)
\((|z_1|^2 - 9)(1 - |z_2|^2) = 0\)
Since it is given that \(|z_2| \neq 1$, we must have \(1 - |z_2|^2 \neq 0\).
Therefore:
\(|z_1|^2 - 9 = 0 \implies |z_1|^2 = 9 \implies |z_1| = 3\)
In simple words: Square both sides and use the property \(|z|^2 = z\bar{z}\) to expand and factor out terms, eliminating \(z_2\) to find \(|z_1|\).

Exam Tip: Memorize the key algebraic identity \(|z|^2 = z\bar{z}\) as it simplifies modulus equations significantly.

Question 24. Find the square root of \(-3 + 4i\) and verify your answer.
Answer: Let \(\sqrt{-3 + 4i} = x + iy\), where \(x, y \in \mathbb{R}\).
Squaring both sides:
\(-3 + 4i = (x + iy)^2\)
\(-3 + 4i = x^2 - y^2 + 2xyi\)
Equating real and imaginary parts:
\(x^2 - y^2 = -3\) --- (1)
\(2xy = 4 \implies xy = 2 \implies y = \frac{2}{x}\) --- (2)
Using the modulus formula \(|x^2 + y^2| = |-3 + 4i|\):
\(x^2 + y^2 = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\) --- (3)
Adding equations (1) and (3):
\(2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1\)
If \(x = 1\), from equation (2), \(y = \frac{2}{1} = 2\).
If \(x = -1\), from equation (2), \(y = \frac{2}{-1} = -2\).
Thus, the square roots are \(\pm(1 + 2i)\).
Verification:
\((1 + 2i)^2 = 1^2 + (2i)^2 + 2(1)(2i) = 1 - 4 + 4i = -3 + 4i\). Verified.
In simple words: Assume the square root is \(x + iy\), equate real and imaginary parts after squaring, use the modulus property to find \(x^2 + y^2\), and solve simultaneously.

Exam Tip: Always verify your calculated square root by squaring it back to check if it matches the original complex number.

Question 25. If \(x = -1 + i\) then find the value of \(x^4 + 4x^3 + 4x^2 + 2\)
Answer: Given \(x = -1 + i\).
Isolate the imaginary term and square both sides:
\(x + 1 = i\)
\((x + 1)^2 = i^2\)
\(x^2 + 2x + 1 = -1\)
\(x^2 + 2x + 2 = 0\) --- (1)
Now, we need to find the value of \(x^4 + 4x^3 + 4x^2 + 2\).
We can divide the polynomial \(x^4 + 4x^3 + 4x^2 + 2\) by the quadratic expression \(x^2 + 2x + 2\):
\(x^4 + 4x^3 + 4x^2 + 2 = (x^2 + 2x + 2)(x^2 + 2x) + 2\)
Since \(x^2 + 2x + 2 = 0\):
\(= (0)(x^2 + 2x) + 2\)
\(= 0 + 2 = 2\)
Wait, let us re-verify via long division or algebraic adjustment:
\(x^4 + 4x^3 + 4x^2 + 2 = (x^2 + 2x + 2)(x^2 + 2x) - 4\)? Let us check carefully.
Let us expand \((x^2 + 2x + 2)(x^2 + 2x) = x^4 + 2x^3 + 2x^3 + 4x^2 + 2x^2 + 4x = x^4 + 4x^3 + 6x^2 + 4x\).
So, \(x^4 + 4x^3 + 4x^2 + 2 = (x^2 + 2x + 2)(x^2 + 2x) - 2x^2 - 4x + 2\).
Since \(x^2 + 2x = -2\), substitute this:
\(= 0 - 2(x^2 + 2x) + 2 = -2(-2) + 2 = 4 + 2 = 6\).
In simple words: Form a quadratic equation from the given complex number and use polynomial division or substitution to evaluate the large expression.

Exam Tip: Reducing the degree of the polynomial using the derived quadratic equation is much faster and less error-prone than substituting \(x = -1 + i\) directly.