Complex Numbers Ques 61
Match the conditions/expressions in Column I with statement in Column II $(z \neq 0$ is a complex number $)$
| Column I | Column II | ||
|---|---|---|---|
| A. | $\operatorname{Re}(z)=0$ | p. | $\operatorname{Re}\left(z^{2}\right)=0$ |
| B. | $\arg (z)=\frac{\pi}{4}$ | q. | $\operatorname{Im}\left(z^{2}\right)=0$ |
| r. | $\operatorname{Re}\left(z^{2}\right)=\operatorname{Im}\left(z^{2}\right)$ |
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Answer:
Correct Answer: 61.$A \rightarrow q ; B \rightarrow p$
Solution:
Formula:
Properties of principal argument(Arg):
- Let $z=a+i b$.
Given, $\operatorname{Re}(z)=0 \Rightarrow a=0$
Then, $z=i b \Rightarrow z^{2}=-b^{2}$ or $\operatorname{lm}\left(z^{2}\right)=0$
Therefore, $A \rightarrow q$
Also, given, $\arg (z)=\frac{\pi}{4}$.
Let
$ z=r \quad (\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}) $
Then,
$ \begin{aligned} z^{2} & =r^{2} (\cos ^{2} \frac{\pi}{4}-\sin ^{2} \frac{\pi}{4})+2 i r^{2} \cos \frac{\pi}{4} \sin \frac{\pi}{4} \\ & =i r^{2} \sin \pi / 2=i r^{2} \end{aligned} $
Therefore, $\operatorname{Re}\left(z^{2}\right)=0 \Rightarrow B \rightarrow p$.
$\Rightarrow$ $a=b=2-\sqrt{3}$
$[\because a, b \leftarrow(0,1)]$
BETA
