Complex Numbers - Algebra PYQ | Mathematics

If \(|z-2|=|z-1|\), where \(z\) is a complex number, then locus \(z\) is a straight line

If \({\left( {{{1 + i} \over {1 - i}}} \right)^m} = 1\), then m cannot be equal to

\((\sin \theta-i \cos \theta)^3\) is equal to

Real part of \((\cos 4+i \sin 4+1)^{2020}\) is

Let \(Z_1, Z_2\) and \(Z_3\) be three non zero complex numbers such that \(a=\left|Z_1\right|, b=\left|Z_2\right|\) and \(c=\left|Z_3\right|\), if the determinant \(\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0\), then

If \(\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2\), where \(z_1\) and \(z_2\) are two complex numbers, then

A real value of \(x\) will satisfy the equation, \(\left(\frac{3-4 i x}{3+4 i x}\right)=\alpha-i \beta,(\alpha, \beta\) are real \()\), if

\(\left(\frac{\sqrt{6}-\sqrt{2}}{4}+\frac{\sqrt{6}+\sqrt{2}}{4} i\right)^{2020}\) is equal to

What is the value of \((1-i \sqrt{3})^9\) is equal to

If \(z_1=2+3 i\) and \(z_2=3+2 i\), where \(i=\sqrt{-1}\), then \(\left[\begin{array}{cc}z_1 & z_2 \\ -\bar{z}_2 & \bar{z}_1\end{array}\right]\left[\begin{array}{cc}\bar{z}_1 & -z_2 \\ \bar{z}_2 & z_1\end{array}\right]\) is equal to

If \(a > 0\) and \(z=x+i y\), then \(\log _{\cos ^2 \theta}|z-a|>\log _{\cos ^2 \theta}|z-a i|,(\theta \in R)\) implies

The radius of the circle represented by \((1+i)(1+3i)(1+7i)=x+iy\) is \((i=\sqrt{-1})\).

If one root of the equation \(i x^2-2(i+1) x+(2-i)=0\) is \((2-i)\), then the other root is

If \(1, \alpha_1, \alpha_2, \alpha_3\) and \(\alpha_4\) are the roots of \(z^5-1=0\) and \(\omega\) is a cube root of units, then \((\omega-1)\left(\omega-\alpha_1\right)\left(\omega-\alpha_2\right)\left(\omega-\alpha_3\right)\left(\omega-\alpha_4\right)+\omega\) is equal to

Multiplicative inverse of the complex number \((\sin \theta, \cos \theta)\) is

$$\sum_\limits{k=0}^{440} i^k=x+i y \Rightarrow x^{100}+x^{99} y+x^{242} y^2+x^{97} y^3=$$

If \(e^{i \theta}=\operatorname{cis} \theta\), then \(\sum_\limits{n=0}^{\infty} \frac{\cos (n \theta)}{2^n}=\)

$$i z^3+z^2-z+i=0 \Rightarrow|z|=$$

If \(\frac{x-1}{3+i}+\frac{y-1}{3-i}=i\), then the true statement among the following is

The number of integer solutions of the equation \(|1-i|^x=2^x\) is