MCQ
If z is any complex number, then $\frac{z-\bar{z}}{2 i}$ is
- Aeither 0 or purely imaginary
- Bpurely imaginary
- Cpurely real
- Deither 0 or purely real
✓
Answer
(c) purely real
Explanation: Let $z = x + i y$
Then $\bar{z}= x - iy$
$\therefore z-\overline{z}=(x+iy)-(x-iy)=2 iy$
Now $\frac{z-\bar{z}}{2 i}=y$
Hence $\frac{z-\bar{z}}{2 i}$ is purely real.
Explanation: Let $z = x + i y$
Then $\bar{z}= x - iy$
$\therefore z-\overline{z}=(x+iy)-(x-iy)=2 iy$
Now $\frac{z-\bar{z}}{2 i}=y$
Hence $\frac{z-\bar{z}}{2 i}$ is purely real.
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