JEE MAIN Complex Number Previous Year Questions (PYQs) – Page 1 of 14
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The region represented by {z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1} is also given by the inequality :{z = x + iy $ \in $ C : |z| – Re(z) $ \le $ 1}
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Solution
Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation
$(\overline{z})^2 + |z| = 0,\; z \in \mathbb{C}$.
Then $4(\alpha^2 + \beta^2)$ is equal to:
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Solution
Let $z = \left(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\right)^5 + \left(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right)^5.$
If $R(z)$ and $I(z)$ respectively denote the real and imaginary parts of $z$, then :
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If $z = x+iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5i|=0$, then :
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The number of complex numbers $z$ satisfying $|z|=1$ and $\left|\dfrac{z}{\overline{z}}+\dfrac{\overline{z}}{z}\right|=1$ is:
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Solution
Given $|z|=1 \Rightarrow z=e^{i\theta}$
$\frac{z}{\overline{z}}=e^{2i\theta},\quad \frac{\overline{z}}{z}=e^{-2i\theta}$
So,
$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right| =|e^{2i\theta}+e^{-2i\theta}|=|2\cos 2\theta|$
Given $|2\cos 2\theta|=1 \Rightarrow |\cos 2\theta|=\frac{1}{2}$
Now in $[0,2\pi)$:
$2\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
⇒ $\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$
But $z=e^{i\theta}$ is periodic with $2\pi$, so each valid $\theta$ gives **2 distinct $z$** over full cycle of $2\theta$
Hence total solutions $=4\times 2=8$
$\frac{z}{\overline{z}}=e^{2i\theta},\quad \frac{\overline{z}}{z}=e^{-2i\theta}$
So,
$\left|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right| =|e^{2i\theta}+e^{-2i\theta}|=|2\cos 2\theta|$
Given $|2\cos 2\theta|=1 \Rightarrow |\cos 2\theta|=\frac{1}{2}$
Now in $[0,2\pi)$:
$2\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
⇒ $\theta=\frac{\pi}{6},\frac{\pi}{3},\frac{2\pi}{3},\frac{5\pi}{6}$
But $z=e^{i\theta}$ is periodic with $2\pi$, so each valid $\theta$ gives **2 distinct $z$** over full cycle of $2\theta$
Hence total solutions $=4\times 2=8$
Let $z$ be the complex number satisfying $|z - 5| \le 3$ and having maximum positive principal argument. Then $34\left|\frac{5z - 12}{5iz + 16}\right|^2$ is equal to:
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Solution
$|z - 5| \le 3$
For $\arg(z)$ to be maximum, $z$ lies at point $P$
$z = (4, 4)$
$= \left(4\left(\frac{4}{5}\right),; 4\left(\frac{3}{5}\right)\right)$
$= \left(\frac{16}{5},; \frac{12}{5}\right)$
Now
$34\left|\frac{5z - 12}{5iz + 16}\right|^2$
$= 34\left|\frac{(16 + 12i) - 12}{(16i + 12i^2) - 12i^2}\right|^2$
$= 34\left|\frac{4 + 12i}{16i + 4}\right|^2$
$= 34\left(\frac{16 + 144}{256 + 16}\right)$
$= 34\left(\frac{160}{272}\right) = 20$
Let C be the set of all complex numbers. Let ${S_1} = \{ z \in C||z - 3 - 2i{|^2} = 8\} $ ${S_2} = \{ z \in C|{\mathop{\rm Re}\nolimits} (z) \ge 5\} $ and ${S_3} = \{ z \in C||z - \overline z | \ge 8\} $. Then the number of elements in ${S_1} \cap {S_2} \cap {S_3}$ is equal to :
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Let z $ \in $ C with Im(z) = 10 and it satisfies ${{2z - n} \over {2z + n}}$ = 2i - 1 for some natural number n. Then :
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Let the product of $\omega_1=(8+i)\sin\theta+(7+4i)\cos\theta$ and $\omega_2=(1+8i)\sin\theta+(4+7i)\cos\theta$ be $\alpha+i\beta$, where $i=\sqrt{-1}$. Let $p$ and $q$ be the maximum and the minimum values of $\alpha+\beta$ respectively. Then $p+q$ is equal to:
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Let $a,b$ be two real numbers such that $ab<0$. If the complex number $\dfrac{1+ai}{\,b+i\,}$ is of unit modulus and $a+ib$ lies on the circle $|z-1|=|2z|$, then a possible value of $\dfrac{1+[a]}{4b}$, where $[\,\cdot\,]$ is the greatest integer function, is:
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Solution
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