If you're preparing for WBJEE or JEE, solving Previous Year Questions (PYQs) is one of the most effective ways to strengthen your preparation. Among all mathematics chapters, Complex Numbers is a high-weightage topic that frequently appears in the WBJEE examination and requires a clear understanding of concepts along with strong problem-solving skills.
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In this post, you'll find year-wise solved WBJEE Complex Numbers questions from 2025 to 2008, arranged in reverse chronological order for easy practice. Each question is accompanied by a detailed solution, helping you understand the underlying concept, improve accuracy, and identify recurring exam patterns.
Whether you're revising the chapter or practicing before the exam, this collection of WBJEE Complex Numbers PYQs (2025–2008) will help you build confidence and maximize your score.
1
WBJEE 2025: Complex Numbers Problems
Detailed Solution:
Since $\frac{2z_1}{3z_2}$ is purely imaginary, $\frac{z_1}{z_2} = ik$ for some real number $k \neq 0$.
Dividing the numerator and denominator by $z_2$:
$$\left| \frac{z_1/z_2 - 1}{z_1/z_2 + 1} \right| = \left| \frac{ik - 1}{ik + 1} \right| = \frac{|ik - 1|}{|ik + 1|} = \frac{\sqrt{k^2 + 1}}{\sqrt{k^2 + 1}} = 1$$
Detailed Solution:
Using the algebraic identity $|a + b\omega + c\omega^2|^2 = \frac{1}{2} \left[ (a-b)^2 + (b-c)^2 + (c-a)^2 \right]$:
Since $a, b, c$ are distinct non-zero integers, the smallest possible squared differences between three distinct integers (for example, $1, 2, 3$) are $1^2, 1^2,$ and $2^2$:
$$\text{Minimum value} = \frac{1}{2} \left[ 1^2 + 1^2 + 2^2 \right] = \frac{1}{2} (1 + 1 + 4) = 3$$
WBJEE 2024: Complex Numbers Problems
Detailed Solution:
For three points $z_1, z_2,$ and origin $0$ to form an equilateral triangle, the condition is:
$$z_1^2 + z_2^2 + 0^2 = z_1 z_2 + z_2 \cdot 0 + 0 \cdot z_1 \implies z_1^2 + z_2^2 = z_1 z_2$$
Rewriting the left side as $(z_1 + z_2)^2 - 2z_1 z_2$, we get:
$$(z_1 + z_2)^2 = 3z_1 z_2$$
From the quadratic equation $z^2 + az + b = 0$, by Vieta's formulas:
$$z_1 + z_2 = -a \quad \text{and} \quad z_1 z_2 = b$$
Substituting these into the identity:
$$(-a)^2 = 3b \implies a^2 = 3b$$
Detailed Solution:
Let $x = e^{i\theta} = \cos\theta + i\sin\theta$. Substituting $x$ into the polynomial equation:
$$a_0 e^{in\theta} + a_1 e^{i(n-1)\theta} + a_2 e^{i(n-2)\theta} + \dots + a_n = 0$$
Multiplying the entire equation by $e^{-in\theta}$:
$$a_0 + a_1 e^{-i\theta} + a_2 e^{-i2\theta} + \dots + a_n e^{-in\theta} = 0$$
Using Euler's identity $e^{-ik\theta} = \cos(k\theta) - i\sin(k\theta)$:
$$a_0 + \sum_{k=1}^{n} a_k \left( \cos k\theta - i\sin k\theta \right) = 0$$
Equating the imaginary part of the complex number to zero gives:
$$-\left( a_1 \sin\theta + a_2 \sin 2\theta + \dots + a_n \sin n\theta \right) = 0$$
$$\implies a_1 \sin\theta + a_2 \sin 2\theta + \dots + a_n \sin n\theta = 0$$
WBJEE 2023: Complex Numbers Problems
Detailed Solution:
Using the property $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$:
• Factor 1: $(1 - \omega + \omega^2) = (1 + \omega^2) - \omega = -\omega - \omega = -2\omega$
• Factor 2: $(1 - \omega^2 + \omega^4) = (1 - \omega^2 + \omega) = (1 + \omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$
• The product of the first pair: $(-2\omega)(-2\omega^2) = 4\omega^3 = 4 = 2^2$
Since the sequence alternates between these two terms for $2n$ total factors (forming $n$ pairs):
$$\text{Product} = \underbrace{(2^2) \times (2^2) \times \dots \times (2^2)}_{n \text{ times}} = (2^2)^n = 2^{2n}$$
Detailed Solution:
Let $w = \frac{z - 2}{z} = 1 - \frac{2}{z}$.
Rearranging for $z$, we get $\frac{2}{z} = 1 - w \implies z = \frac{2}{1 - w}$.
Substitute $z$ into the given condition $|z - 1| = 1$:
$$\left| \frac{2}{1 - w} - 1 \right| = 1 \implies \left| \frac{2 - (1 - w)}{1 - w} \right| = 1 \implies |1 + w| = |1 - w|$$
Squaring both sides with $w = u + iv$:
$$(u + 1)^2 + v^2 = (1 - u)^2 + v^2 \implies u^2 + 2u + 1 = 1 - 2u + u^2 \implies 4u = 0 \implies u = 0$$
Since the real part $u = 0$, $w$ lies entirely on the **imaginary axis**.
WBJEE 2022: Complex Numbers Problems
Detailed Solution:
Since the base of the logarithm is $\frac{1}{2} < 1$, applying the anti-logarithm reverses the inequality direction:
$$\log_{1/2} |z - 1| > \log_{1/2} |z - i| \implies |z - 1| < |z - i|$$
Let $z = x + iy$. Squaring both sides:
$$|x + iy - 1|^2 < |x + iy - i|^2$$
$$(x - 1)^2 + y^2 < x^2 + (y - 1)^2$$
$$x^2 - 2x + 1 + y^2 < x^2 + y^2 - 2y + 1$$
$$-2x < -2y \implies x > y \text{ (or } y < x \text{)}$$
Hence, the region satisfying the condition is $x < y$ when framed as $|z - 1| < |z - i|$.
Detailed Solution:
Substitute $z = x + iy$ into the expression:
$$\frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy}$$
Multiply numerator and denominator by the conjugate of the denominator, $(x + 1) - iy$:
$$\frac{[(x - 1) + iy][(x + 1) - iy]}{(x + 1)^2 + y^2} = \frac{(x^2 - 1 + y^2) + i(2y)}{(x + 1)^2 + y^2}$$
Setting the real part to zero:
$$\frac{x^2 + y^2 - 1}{(x + 1)^2 + y^2} = 0 \implies x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1$$
This represents a circle centered at the origin with radius $1$. Since $z \neq -1$, the point $(-1,0)$ is excluded.
WBJEE 2021: Complex Numbers Problems
Detailed Solution:
Using the property of arguments $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$:
• $\arg(1 + i\sqrt{3}) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
• $\arg(1 + i) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$
$$\text{Principal Argument} = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$$
Detailed Solution:
To determine the nature of $w$, consider its real part or conjugate properties:
$$\bar{w} = \overline{\left(\frac{z - 1}{z + 1}\right)} = \frac{\bar{z} - 1}{\bar{z} + 1}$$
Since $|z| = 1 \implies z\bar{z} = 1 \implies \bar{z} = \frac{1}{z}$:
$$\bar{w} = \frac{\frac{1}{z} - 1}{\frac{1}{z} + 1} = \frac{1 - z}{1 + z} = -\left(\frac{z - 1}{z + 1}\right) = -w$$
Since $\bar{w} = -w$, $w + \bar{w} = 0$, which means $\operatorname{Re}(w) = 0$.
Therefore, $w$ is purely imaginary, and its locus is the **imaginary axis**.
WBJEE 2020: Complex Numbers Problems
Detailed Solution:
Using the identity $1 + \omega + \omega^2 = 0$:
• $1 + \omega^2 = -\omega \implies (1 - \omega + \omega^2)^5 = (-\omega - \omega)^5 = (-2\omega)^5 = -32\omega^5 = -32\omega^2$
• $1 + \omega = -\omega^2 \implies (1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5 = -32\omega^{10} = -32\omega$
Summing both parts:
$$-32\omega^2 - 32\omega = -32(\omega^2 + \omega) = -32(-1) = 32$$
Detailed Solution:
The equation $|z - z_1| = |z - z_2|$ represents the perpendicular bisector of the line segment joining $z_1(2, 3)$ and $z_2(3, 2)$.
Squaring both sides:
$$(x - 2)^2 + (y - 3)^2 = (x - 3)^2 + (y - 2)^2$$
$$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 6x + 9 + y^2 - 4y + 4$$
$$-4x - 6y = -6x - 4y \implies 2x - 2y = 0 \implies x - y = 0$$
WBJEE 2019: Complex Numbers Problems
Detailed Solution:
The roots of $z^2 + z + 1 = 0$ are the non-real cube roots of unity, $\omega$ and $\omega^2$.
Using the properties $\omega^3 = 1$ and $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$:
• For $k = 1$: $\left(\omega + \frac{1}{\omega}\right)^2 = (-1)^2 = 1$
• For $k = 2$: $\left(\omega^2 + \frac{1}{\omega^2}\right)^2 = \left(\omega^2 + \omega\right)^2 = (-1)^2 = 1$
• For $k = 3$: $\left(\omega^3 + \frac{1}{\omega^3}\right)^2 = (1 + 1)^2 = 2^2 = 4$
Because the values repeat every $3$ terms, the $6$ terms consist of two identical sets of $(1 + 1 + 4)$:
$$\text{Sum} = 2 \times (1 + 1 + 4) = 2 \times 6 = 12$$
Detailed Solution:
Squaring both sides of $|z_1 - z_2| = |z_1 + z_2|$:
$$(z_1 - z_2)(\bar{z}_1 - \bar{z}_2) = (z_1 + z_2)(\bar{z}_1 + \bar{z}_2)$$
$$|z_1|^2 - z_1\bar{z}_2 - z_2\bar{z}_1 + |z_2|^2 = |z_1|^2 + z_1\bar{z}_2 + z_2\bar{z}_1 + |z_2|^2$$
$$-2(z_1\bar{z}_2 + z_2\bar{z}_1) = 2(z_1\bar{z}_2 + z_2\bar{z}_1) \implies 4 \operatorname{Re}(z_1\bar{z}_2) = 0$$
Divide by $|z_2|^2$:
$$\operatorname{Re}\left(\frac{z_1}{z_2}\right) = 0$$
Since the real part is zero, $\frac{z_1}{z_2}$ is **purely imaginary**.
WBJEE 2018: Complex Numbers Problems
Detailed Solution:
We know that $1 + \omega + \omega^2 = 0 \implies 1 + \omega = -\omega^2$.
Substituting this into the expression:
$$(1 + \omega - \omega^2)^7 = (-\omega^2 - \omega^2)^7 = (-2\omega^2)^7$$
$$= (-2)^7 \cdot (\omega^2)^7 = -128 \cdot \omega^{14}$$
Since $\omega^3 = 1$, we simplify $\omega^{14} = (\omega^3)^4 \cdot \omega^2 = \omega^2$.
Therefore, the value is **$-128\omega^2$**.
Detailed Solution:
Using the reverse triangle inequality $|z_1 + z_2| \ge ||z_1| - |z_2||$:
$$\left|z + \frac{1}{2}\right| \ge ||z| - \left|\frac{1}{2}\right||$$
Given $|z| \ge 2$, we have $|z| - \frac{1}{2} \ge 2 - \frac{1}{2} = \frac{3}{2}$.
Hence, the minimum value is **$3/2$** (attained when $z = -2$).
WBJEE 2017: Complex Numbers Problems
Detailed Solution:
Let $\theta = \frac{\pi}{8}$. Convert sine to cosine: $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ and $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$.
Let $\alpha = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$. The fraction becomes:
$$\frac{1 + \cos\alpha + i\sin\alpha}{1 + \cos\alpha - i\sin\alpha} = \frac{2\cos^2\frac{\alpha}{2} + 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2} - 2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} = \frac{\cos\frac{\alpha}{2} + i\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2} - i\sin\frac{\alpha}{2}} = e^{i\alpha}$$
Raising this expression to the 8th power:
$$(e^{i\alpha})^8 = e^{i8\alpha} = \cos(8\alpha) + i\sin(8\alpha)$$
Since $\alpha = \frac{3\pi}{8}$, we have $8\alpha = 3\pi$.
$$\cos(3\pi) + i\sin(3\pi) = -1 + 0 = -1$$
Detailed Solution:
Using the triangle inequality $|z_1 - z_2| \ge ||z_1| - |z_2||$:
$$\left|z - \frac{1}{z}\right| \ge ||z| - \frac{1}{|z|}||$$
Given $\left|z - \frac{1}{z}\right| = 2$, we get:
$$||z| - \frac{1}{|z|}|| \le 2 \implies -2 \le |z| - \frac{1}{|z|} \le 2$$
Taking the upper inequality $|z| - \frac{1}{|z|} \le 2$ and substituting $r = |z|$ ($r > 0$):
$$r^2 - 2r - 1 \le 0$$
Solving the quadratic equation $r^2 - 2r - 1 = 0$ gives $r = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.
Since $r > 0$, the roots bound $r$ such that $0 < r \le \sqrt{2} + 1$.
Hence, the maximum value of $|z|$ is **$\sqrt{2} + 1$**.
WBJEE 2016: Complex Numbers Problems
Detailed Solution:
Using $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$:
• 1st factor: $1 - \omega + \omega^2 = -\omega - \omega = -2\omega$
• 2nd factor: $1 - \omega^2 + \omega^4 = 1 - \omega^2 + \omega = -\omega^2 - \omega^2 = -2\omega^2$
The product of the first two factors is $(-2\omega)(-2\omega^2) = 4\omega^3 = 4 = 2^2$.
Since powers of $\omega$ repeat with period $3$, every pair of successive factors evaluates to $2^2 = 4$.
For $2n$ factors, there are $n$ such pairs:
$$\text{Product} = (2^2)^n = 2^{2n}$$
Detailed Solution:
The condition $|z - (-1)| = |z - 1|$ defines points $z$ equidistant from $(-1, 0)$ and $(1, 0)$.
Squaring both sides:
$$(x + 1)^2 + y^2 = (x - 1)^2 + y^2$$
$$x^2 + 2x + 1 + y^2 = x^2 - 2x + 1 + y^2$$
$$4x = 0 \implies x = 0$$
$x = 0$ represents the **y-axis** (purely imaginary axis).
WBJEE 2015: Complex Numbers Problems
Detailed Solution:
The roots of $x^2 - x + 1 = 0$ are $\alpha = -\omega$ and $\beta = -\omega^2$, where $\omega$ is a non-real cube root of unity.
Substituting these into the expression:
$$\alpha^{2015} + \beta^{2015} = (-\omega)^{2015} + (-\omega^2)^{2015}$$
Since $2015$ is odd, $(-1)^{2015} = -1$:
$$= -(\omega^{2015} + \omega^{4030})$$
Using $\omega^3 = 1$:
• $2015 = 3 \times 671 + 2 \implies \omega^{2015} = \omega^2$
• $4030 = 3 \times 1343 + 1 \implies \omega^{4030} = \omega$
$$\text{Sum} = -(\omega^2 + \omega) = -(-1) = 1$$
Detailed Solution:
The equation is of the form $|z - z_1| + |z - z_2| = 2a$.
Here, $z_1 = 4$, $z_2 = -4$, and the constant sum is $2a = 10$.
The distance between the fixed points (foci) is $|z_1 - z_2| = |4 - (-4)| = 8$.
Since $2a = 10 > 8$ (the distance between foci), the sum of distances from $z$ to two fixed points is a constant greater than the distance between them.
This is the standard definition of an **ellipse**.
WBJEE 2014: Complex Numbers Problems
Detailed Solution:
Using the relations $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$:
• $1 - \omega + \omega^2 = -\omega - \omega = -2\omega$
• $1 - \omega^2 + \omega^4 = 1 - \omega^2 + \omega = -\omega^2 - \omega^2 = -2\omega^2$
• $1 - \omega^4 + \omega^8 = 1 - \omega + \omega^2 = -2\omega$
• $1 - \omega^8 + \omega^{16} = 1 - \omega^2 + \omega = -2\omega^2$
Multiplying all four factors together:
$$(-2\omega)(-2\omega^2)(-2\omega)(-2\omega^2) = (4\omega^3)(4\omega^3) = 4 \times 4 = 16$$
Detailed Solution:
Substitute $z = x + iy$ into $\frac{z - 1}{z + 1}$:
$$\frac{(x - 1) + iy}{(x + 1) + iy} = \frac{[(x - 1) + iy][(x + 1) - iy]}{(x + 1)^2 + y^2} = \frac{(x^2 + y^2 - 1) + i(2y)}{(x + 1)^2 + y^2}$$
Since $\operatorname{Amp}\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{4}$, we have $\tan\left(\frac{\pi}{4}\right) = \frac{\text{Im}}{\text{Re}} = 1$:
$$\frac{2y}{x^2 + y^2 - 1} = 1 \implies x^2 + y^2 - 2y - 1 = 0$$
WBJEE 2013: Complex Numbers Problems
Detailed Solution:
Using the reverse triangle inequality:
$$\left|z + \frac{1}{2}\right| \ge ||z| - \left|-\frac{1}{2}\right|| = ||z| - \frac{1}{2}|$$
Given that $|z| \ge 2$ and the function $f(x) = x - \frac{1}{2}$ is monotonically increasing for $x \ge 2$:
$$\text{Minimum value} = 2 - \frac{1}{2} = \frac{3}{2}$$
Detailed Solution:
We know that for any complex number $z$, $\bar{z} = \frac{1}{z}$ when $|z| = 1$.
Since $|z_1| = |z_2| = |z_3| = 1$, we have $\frac{1}{z_1} = \bar{z_1}$, $\frac{1}{z_2} = \bar{z_2}$, and $\frac{1}{z_3} = \bar{z_3}$.
Given $\left|\frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3}\right| = 1$:
$$\implies |\bar{z_1} + \bar{z_2} + \bar{z_3}| = 1$$
Using the property $|\bar{z}| = |z|$:
$$\implies \overline{|z_1 + z_2 + z_3|} = |z_1 + z_2 + z_3| = 1$$
WBJEE 2012: Complex Numbers Problems
Detailed Solution:
Using the property $1 + \omega + \omega^2 = 0$, we get $1 + \omega = -\omega^2$.
Substituting this into the expression:
$$\left(1 + \omega - \omega^2\right)^7 = \left(-\omega^2 - \omega^2\right)^7 = \left(-2\omega^2\right)^7$$
$$= (-2)^7 \cdot (\omega^2)^7 = -128 \cdot \omega^{14}$$
Since $\omega^3 = 1$, we simplify $\omega^{14} = (\omega^3)^4 \cdot \omega^2 = \omega^2$:
$$= -128\omega^2$$
Detailed Solution:
From the given equation:
$$\left|\frac{z - i}{z + i}\right| = 1 \implies |z - i| = |z + i|$$
This represents the set of points $z$ that are equidistant from the points $i = (0, 1)$ and $-i = (0, -1)$.
The perpendicular bisector of the line segment joining $(0, 1)$ and $(0, -1)$ is the x-axis ($y = 0$).
Alternatively, setting $z = x + iy$:
$$x^2 + (y - 1)^2 = x^2 + (y + 1)^2 \implies -2y = 2y \implies y = 0$$
Thus, the locus is the **real axis**.
WBJEE 2011: Complex Numbers Problems
Detailed Solution:
We know that $\operatorname{Amp}\left(\frac{z_1}{z_2}\right) = \operatorname{Amp}(z_1) - \operatorname{Amp}(z_2)$.
Let $z_1 = 1 + i\sqrt{3}$ and $z_2 = 1 - i\sqrt{3}$.
• $\operatorname{Amp}(z_1) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
• $\operatorname{Amp}(z_2) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$
Therefore:
$$\operatorname{Amp}(z) = \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) = \frac{2\pi}{3}$$
Detailed Solution:
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$$\begin{vmatrix} 1+\omega+\omega^2 & \omega & \omega^2 \\ 1+\omega+\omega^2 & \omega^2 & 1 \\ 1+\omega+\omega^2 & 1 & \omega \end{vmatrix}$$
Since $1 + \omega + \omega^2 = 0$, the first column becomes entirely zero:
$$\begin{vmatrix} 0 & \omega & \omega^2 \\ 0 & \omega^2 & 1 \\ 0 & 1 & \omega \end{vmatrix} = 0$$
WBJEE 2010: Complex Numbers Problems
Detailed Solution:
Solving the quadratic equation $x^2 - (2\cos\theta)x + 1 = 0$ gives:
$$x = \cos\theta \pm i\sin\theta = e^{\pm i\theta}$$
Using De Moivre's Theorem:
$$x^n = e^{\pm in\theta} = \cos(n\theta) \pm i\sin(n\theta)$$
$$\frac{1}{x^n} = e^{\mp in\theta} = \cos(n\theta) \mp i\sin(n\theta)$$
Adding the two terms yields:
$$x^n + \frac{1}{x^n} = 2\cos(n\theta)$$
Detailed Solution:
Using the relation $1 + \omega + \omega^2 = 0$:
1. $1 + \omega = -\omega^2 \implies 1 + \omega - \omega^2 = -2\omega^2$
2. $1 + \omega^2 = -\omega \implies 1 - \omega + \omega^2 = -2\omega$
Substituting these into the expression:
$$(-2\omega^2)^3 - (-2\omega)^3 = -8\omega^6 - (-8\omega^3)$$
Since $\omega^3 = 1$ and $\omega^6 = 1$:
$$= -8(1) + 8(1) = 0$$
WBJEE 2009: Complex Numbers Problems
Detailed Solution:
Using $\omega^3 = 1$ to simplify the exponents:
• $1 + \omega^4 = 1 + \omega$
• $1 + \omega^8 = 1 + \omega^2$
Thus, the expression becomes:
$$(1 + \omega)(1 + \omega^2)(1 + \omega)(1 + \omega^2) = [(1 + \omega)(1 + \omega^2)]^2$$
Using $1 + \omega + \omega^2 = 0 \implies 1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$:
$$[(-\omega^2)(-\omega)]^2 = [\omega^3]^2 = 1^2 = 1$$
Detailed Solution:
Substituting $z = x + iy$ into the given equation $|z - 1| = |z + i|$:
$$|(x - 1) + iy| = |x + i(y + 1)|$$
Squaring both sides:
$$(x - 1)^2 + y^2 = x^2 + (y + 1)^2$$
$$x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2y + 1$$
$$-2x = 2y \implies y = -x$$
This is a line passing through the origin of the form $y = mx$ with slope $m = -1$.
WBJEE 2008: Complex Numbers Problems
Detailed Solution:
Using the property $\arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2)$:
• For $z_1 = 1 + i\sqrt{3}$, $\arg(z_1) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$
• For $z_2 = \sqrt{3} + i$, $\arg(z_2) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$
Therefore, $\arg(z) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Detailed Solution:
Using the identity $1 + \omega + \omega^2 = 0$:
1. $1 + \omega^2 = -\omega \implies 1 - \omega + \omega^2 = -2\omega$
2. $1 + \omega = -\omega^2 \implies 1 + \omega - \omega^2 = -2\omega^2$
Substituting these into the expression:
$$(-2\omega)^5 + (-2\omega^2)^5 = -32\omega^5 - 32\omega^{10}$$
Since $\omega^3 = 1$, we have $\omega^5 = \omega^2$ and $\omega^{10} = \omega$:
$$= -32(\omega^2 + \omega)$$
Since $\omega + \omega^2 = -1$:
$$= -32(-1) = 32$$
