Jamia Millia Islamia Complex Number Previous Year Questions (PYQs) – Page 2 of 3
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Showing 27 questions
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 $z=\tfrac{3}{2}-2i$.
Which of the following expresses the given complex number $(1 - i)^4$
in the form $(a + ib)$?
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Solution
$(1 - i)^4 = (1 - i)^2 \times (1 - i)^2$
$= (1 - 2i + i^2)^2 = (1 - 2i - 1)^2 = (-2i)^2 = -4$
The product of complex numbers $(4,3)$ and $(5,-6)$ is?
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Solution
$(4 + 3i)(5 - 6i) = 20 - 24i + 15i - 18i^2 = 20 - 9i + 18 = 38 - 9i.$
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, $\tan\theta = \dfrac{\text{Imag}}{\text{Real}} = \dfrac{\sqrt{3}+3}{\sqrt{3}-1} = \tan\left(\dfrac{\pi}{3}\right)$.
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.
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.
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:
$(1 + \omega - \omega^{2})(1 - \omega + \omega^{2})
= 1 - \omega^{2} + \omega^{4} - \omega + \omega^{2} - \omega^{3} + \omega^{2} - \omega^{3} + \omega^{4}
= 1 + 3 = 3.$
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 $k = 2$, $x = 1 - 2\omega^{2}$.
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 $=\big||9|-4\big|=5$.
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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