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Jamia Millia Islamia Previous Year Questions (PYQs)
Jamia Millia Islamia Complex Number PYQ
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The value of $\tan 3A - \tan 2A - \tan A$ is equal to:
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Solution
Since $1^\circ = \dfrac{\pi}{180}$ radians and $\sin x$ increases for small $x$ in radians, $\sin(1^\circ) = \sin\left(\dfrac{\pi}{180}\right) \approx 0.01745 < \sin(1 \text{ rad}) \approx 0.841$. Hence, $\sin 1^\circ < \sin 1$.
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Number of unimodular complex numbers which satisfy the locus
$\arg\!\left(\dfrac{z - 1}{z + i}\right) = \dfrac{\pi}{2}$ is —
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Solution
Let $z = x + iy$ and $|z| = 1$. The given condition means $(z - 1)$ is perpendicular to $(z + i)$. \[ (x - 1, y) \cdot (x, y + 1) = 0 \Rightarrow x^2 - x + y^2 + y = 0 \] Since $x^2 + y^2 = 1$ (unimodular), \[ 1 - x + y = 0 \Rightarrow y = x - 1. \] Substitute in $x^2 + y^2 = 1$: \[ x^2 + (x - 1)^2 = 1 \Rightarrow 2x^2 - 2x = 0 \Rightarrow x(x - 1) = 0. \] Hence $x = 0$ or $x = 1$. For $x = 0$, $z = -i$ (denominator $z+i=0$). For $x = 1$, $z = 1$ (numerator $z-1=0$). Both make $\arg$ undefined. Therefore, no unimodular complex number satisfies the condition.
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The tens digit of $1! + 2! + 3! + 4! + \dots + 49!$ is —
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Solution
From $10!$ onward, every factorial ends with at least two zeros. So, only the sum of first nine factorials affects the tens digit. $1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 1 + 2 + 6 + 24 + 120 + 720 + 5040 + 40320 + 362880 = 409113.$ Tens digit = $\boxed{1}.$
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If $\omega$ is a cube root of unity, then
$\left|\begin{array}{ccc}
1 & \omega & \omega^2 \\
1 & \omega^2 & 1 \\
\omega & 1 & \omega^2
\end{array}\right|$
is equal to
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Solution
Use $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. Evaluating the determinant gives value $= -3$.
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If $49^{n}+16n+\lambda$ is divisible by $64$ for all $n\in\mathbb{N}$,
then the least negative value of $\lambda$ is –
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Solution
Solution: Work modulo $64$. $49\equiv -15\pmod{64}$ and $49^{1}\equiv49,\ 49^{2}\equiv33,\ 49^{3}\equiv17,\ 49^{4}\equiv1$; hence $49^{n}$ is periodic with period $4$. Also $16n\equiv 0,16,32,48\ (\bmod\ 64)$ for $n\equiv 0,1,2,3$. For each residue class: $n\equiv0$: $1+0+\lambda\equiv0\Rightarrow \lambda\equiv -1$ $n\equiv1$: $49+16+\lambda\equiv1+\lambda\equiv0$ $n\equiv2$: $33+32+\lambda\equiv1+\lambda\equiv0$ $n\equiv3$: $17+48+\lambda\equiv1+\lambda\equiv0$ All give $\lambda\equiv -1\pmod{64}$. The least negative representative is $\boxed{-1}$.
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Let $z=\sqrt{3}+i$ be a complex number and $\bar z$ its conjugate. The $|\arg z|+|\arg \bar z|$ is equal to
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Solution
$\arg z = \tan^{-1}\!\left(\frac{1}{\sqrt3}\right)=\frac{\pi}{6}$ (I quadrant), $\arg\bar z = -\frac{\pi}{6}$. Sum of absolute values $=\frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3}$.
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The $\dfrac{(\sqrt3+i)^{17}}{(1-i)^{50}}$ is equal to …
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Solution
$\sqrt{3}+i = 2(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}) \Rightarrow (\sqrt{3}+i)^{17} = 2^{17}\left[\cos\tfrac{17\pi}{6} + i\sin\tfrac{17\pi}{6}\right] = 2^{16}(-\sqrt{3}+i).$ $1 - i = \sqrt{2}(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4})) \Rightarrow (1-i)^{50} = 2^{25}\left[\cos(-\tfrac{25\pi}{2}) + i\sin(-\tfrac{25\pi}{2})\right] = -i\,2^{25}.$ $\text{Ratio} = 2^{-9}\dfrac{-\sqrt{3}+i}{-i} = 2^{-9}(-1-\sqrt{3}i).$
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For which value of $x$, $\left(\dfrac{1+i}{1-i}\right)^{x}=1$ ?
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Solution
$\dfrac{1+i}{1-i}=i$. So $i^{x}=1\Rightarrow x\equiv0\pmod4$. Only $68$ is a multiple of $4$.
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If $\omega$ is a cube root of unity, the value of $(1-\omega-\omega^{2})(1+\omega^{3})$ is …
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Solution
$\omega^{3}=1$ and $\omega+\omega^{2}=-1$. So $1-\omega-\omega^{2}=2$ and $1+\omega^{3}=2$; product $=4$.
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If $\left(\dfrac{1+i}{1-i}\right)^{x}=1$, then
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Solution
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Let $z$ be a complex number. Which of the following is a solution of $|z|-z=1+2i$?
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Solution
Let $z=x+iy$. Then $|z|-x-iy=1+2i \Rightarrow |z|-x=1,\; -y=2\Rightarrow y=-2$. $|z|=1+x$, and $(1+x)^{2}=x^{2}+4 \Rightarrow x=\tfrac{3}{2}$. Thus $z=\tfrac{3}{2}-2i$.
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Which of the following expresses the given complex number $(1 - i)^4$
in the form $(a + ib)$?
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Solution
$(1 - i)^4 = (1 - i)^2 \times (1 - i)^2$ $= (1 - 2i + i^2)^2 = (1 - 2i - 1)^2 = (-2i)^2 = -4$
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The product of complex numbers $(4,3)$ and $(5,-6)$ is?
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Solution
$(4 + 3i)(5 - 6i) = 20 - 24i + 15i - 18i^2 = 20 - 9i + 18 = 38 - 9i.$
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The amplitude of $\dfrac{1+i\sqrt{3}}{\sqrt{3}+i}$ is equal to …
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Solution
$\dfrac{1+i\sqrt{3}}{\sqrt{3}+i} = \dfrac{(1+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)} = \dfrac{(\sqrt{3}+3)i + (\sqrt{3}-1)}{4}$. Hence, $\tan\theta = \dfrac{\text{Imag}}{\text{Real}} = \dfrac{\sqrt{3}+3}{\sqrt{3}-1} = \tan\left(\dfrac{\pi}{3}\right)$.
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If $z$ be a complex number and $\bar{z}$ be its conjugate, then the number of solutions of $z^{2} + 2\bar{z} = 0$ is …
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Solution
Let $z = x + iy$, then $\bar{z} = x - iy$. Substitute: $(x + iy)^{2} + 2(x - iy) = 0$. $\Rightarrow x^{2} - y^{2} + 2ixy + 2x - 2iy = 0$. Equating real and imaginary parts: Real: $x^{2} - y^{2} + 2x = 0$ Imag: $2xy - 2y = 0 \Rightarrow y( x - 1 ) = 0$. If $y=0$, then $x^{2} + 2x = 0 \Rightarrow x = 0, -2$. If $x=1$, then $1 - y^{2} + 2 = 0 \Rightarrow y^{2} = 3$. Hence, total 4 solutions.
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If $z$ be a complex number, then one of the solutions of the equation $z^{2} + |z|^{2} = 0$ is …
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Solution
Let $z = x + iy$, then $|z|^{2} = x^{2} + y^{2}$. So, $(x + iy)^{2} + x^{2} + y^{2} = 0$. $\Rightarrow (2x^{2} - y^{2}) + 2ixy = 0 \Rightarrow 2x^{2} - y^{2} = 0, 2xy = 0$. If $x = 0$, then $-y^{2} = 0 \Rightarrow y = 0$. If $y = 0$, then $2x^{2} = 0 \Rightarrow x = 0$. Non-trivial solution: $x = y = 0$. So we take $z = i\sqrt{2}|x|$ form. $z = 2 + 3i$ satisfies the given equation.
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If $\omega$ is a cube root of unity, then the value of $(1 + \omega - \omega^{2})(1 - \omega + \omega^{2})$ is …
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Solution
We know $\omega^{3} = 1$ and $1 + \omega + \omega^{2} = 0$. Expanding: $(1 + \omega - \omega^{2})(1 - \omega + \omega^{2}) = 1 - \omega^{2} + \omega^{4} - \omega + \omega^{2} - \omega^{3} + \omega^{2} - \omega^{3} + \omega^{4} = 1 + 3 = 3.$
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Let cube roots of unity be $1, \omega, \omega^{2}$. Which of the following is a cube root of the equation $(x - 1)^{3} + 8 = 0$?
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Solution
$(x - 1)^{3} = -8 \Rightarrow x - 1 = 2\omega^{k}$ where $k = 0, 1, 2$. So, $x = 1 + 2\omega^{k}$. For $k = 2$, $x = 1 - 2\omega^{2}$.
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If $z$ is any complex number with $|z-3-2i|\le2$, then the minimum of $|\,2z-6+5i\,|$ is:
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Solution
Put $w=z-(3+2i)$ so $|w|\le2$. Then $2z-6+5i=2w+9i$. Set of $2w$ is a disc of radius $4$ centered at $0$, so $2w+9i$ is a disc of radius $4$ centered at $9i$. Minimum modulus $=\big||9|-4\big|=5$.
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For $z\ne0$, the value of $\arg z+\arg\overline{z}$ is:
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Solution
$\arg(\overline{z})=-\arg(z)$ (principal values). Hence sum $=0$.
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The number of complex numbers $Z$ such that
$|Z - 1| = |Z + 1| = |Z - i|$ is:
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Solution
$|Z - 1| = |Z + 1|$ represents the perpendicular bisector of the line joining $(1,0)$ and $(-1,0)$, i.e. the $y$–axis. For points on the $y$–axis, $Z = i y$. Now $|Z - i| = |i y - i| = |i(y - 1)| = |y - 1|$. Also $|Z + 1| = \sqrt{1 + y^2}$. So $\sqrt{1 + y^2} = |y - 1| \Rightarrow y = 0$. Thus only one point satisfies — $Z = 0$.
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If $\omega$ is a cube root of unity and $(1 + \omega)^7 = A + B\omega$, then $A + B$ equals:
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Solution
We know $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. So $1 + \omega = -\omega^2$. Hence $(1 + \omega)^7 = (-\omega^2)^7 = (-1)^7 \omega^{14} = -\omega^{14}$. Now $\omega^{14} = \omega^{3 \times 4 + 2} = \omega^2$. Therefore $(1 + \omega)^7 = -\omega^2 = 1 + \omega$. So comparing with $A + B\omega$, we have $A = 1$, $B = 1$. $\Rightarrow A + B = 2.$
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If $|z_1| = 4,\ |z_2| = 3$, then what is the value of $|z_1 + z_2 + 3 + 4i|$ ?
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Solution
Using triangle inequality,
Hence, value is less than 12.
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If $\omega$ is an imaginary cube root of unity, then $(1 + \omega - \omega^2)^7$ equals
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Solution
$1+\omega+\omega^2=0 \Rightarrow 1+\omega-\omega^2=2\omega.$ $(2\omega)^7 = 128\omega^7 = 128\omega.$
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The complex numbers $\sin x + i\cos 2x$ and $\cos x - i\sin 2x$ are conjugate to each other for
A. $x = n\pi$ B. $x = 0$ C. $x = (n + \tfrac{1}{2})\pi$ D. No value of $x$
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Solution
Conjugates ⇒ $\cos 2x = \sin x \Rightarrow x = (n + \tfrac{1}{2})\pi.$
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The points $z_1, z_2, z_3, z_4$ in the complex plane form a parallelogram if and only if
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Solution
Diagonals bisect each other ⇒ $z_1 + z_3 = z_2 + z_4.$Jamia Millia Islamia
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