Options : $\begin{array}{llll}A & B & C & D\end{array}$
If $Z_1$ and $Z_2$ are conjugate complex numbers. Match the items under the following columns? Options :…
Options : $\begin{array}{llll}A & B & C & D\end{array}$
- $\begin{array}{llll}3 & 1 & 4 & 2\end{array}$
- $\begin{array}{llll}3 & 1 & 2 & 4\end{array}$
- $\begin{array}{llll}1 & 2 & 4 & 3\end{array}$
- $\begin{array}{llll}2 & 4 & 3 & 1\end{array}$
Solution
Let $Z_1=a+i b$ $\Rightarrow \quad\left|Z_1\right|^2=\left(\sqrt{a^2+b^2}\right)^2$ $=a^2+b^2$ $\Rightarrow \quad Z_2=a-i b$ (i) $Z_1 Z_2=(a+i b)(a-i b)$ $=a^2+b^2=\left|Z_1\right|^2$ (ii) $Z_1+Z_2=0$ $a+i b+a-i b=0$ $2 a=0 \Rightarrow a=0$ $\Rightarrow$ imaginary axis. (iii) If $Z_1=a+i b$ $\begin{aligned} & \operatorname{Im}\left(Z_1\right)=b \\ & \operatorname{Im}\left(-Z_2\right)=\operatorname{Im}(-(a-i b))\end{aligned}$ $=\operatorname{Im}(-a+i b)=b$ Hence, $\operatorname{Im}\left(Z_1\right)=\operatorname{Im}\left(-Z_2\right)$ (iv) $\operatorname{Re}\left(Z_1\right)=a$ $\operatorname{Re}\left(Z_2\right)=a$ $\Rightarrow \operatorname{Re}\left(Z_1\right)=\operatorname{Re}\left(Z_2\right)$ $\therefore \mathrm{A}-(3) ; \mathrm{B}-(1) ; \mathrm{C}-(2) ; \mathrm{D}-(4)$.
Asked in: AP EAMCET 2021 (23 Aug Shift 2)