Question
If $\left|z_{1}+z_{2}\right|=\left|z_{1}-z_{2}\right|$ and $\left|z_{2}\right|=\sqrt{3}\left|z_{1}\right|$, then the absolute value of the difference of the arguments of $z_{1}+z_{2}$ and $\mathrm{z}_{1}$ is(a) 0(b) $\frac{\pi}{2}$(c) $\frac{\pi}{3}$(d) $\frac{\pi}{6}$
Step 1
Step 1: Consider the parallelogram on the complex plane formed by vertices at the origin, $z_{1}$, $z_{2}$, and $z_{1}+z_{2}$. Show more…
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If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $\left|z_{1}+z_{2}\right|=\left|z_{1}\right|+\left|z_{2}\right|$, then $\arg z_{1}-\arg z_{2}$ is equal to (A) $-\pi$ (B) $-\frac{\pi}{2}$ (C) $\pi$ (D) $\frac{\pi}{2}$
If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}\left(z_{1}\right)=\left|z_{1}-1\right|, \operatorname{Re}\left(z_{2}\right)=\mid z_{2}-$ 1| and $\arg \left(z_{1}-z_{2}\right)=\frac{\pi}{6}$, then $\operatorname{Im}\left(z_{1}+z_{2}\right)$ is equal to (a) $\frac{2}{\sqrt{3}}$ (b) $2 \sqrt{3}$ (c) $\frac{\sqrt{3}}{2}$ (d) $\frac{1}{\sqrt{3}}$
$\mathrm{A}=\left[\begin{array}{ccc}1+\mathrm{z}^{2} & 1 & 1 \\ 0 & 1+\mathrm{z}^{2} & 1 \\ 0 & 0 & 1+\mathrm{z}^{2}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}1+\mathrm{z}^{-2} & 0 & 0 \\ 1 & 1+\mathrm{z}^{-2} & 0 \\ 1 & 1 & 1+\mathrm{z}^{-2}\end{array}\right]$ where, $|\mathrm{z}|=1$. If $|\mathrm{A}+\mathrm{B}|=0$ then one value of $\arg (\mathrm{z})$ is (a) $\frac{\pi}{4}$ (b) $\frac{\pi}{6}$ (c) $\frac{2 \pi}{3}$ (d) $\frac{3 \pi}{4}$
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