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Question: If \|z\| = 1 and p = \(\frac{z–1}{z + 1}\) (where z ¹ –1) then Re(p) equals –...

If |z| = 1 and p = z–1z+1\frac{z–1}{z + 1} (where z ¹ –1) then Re(p) equals –

A

0

B

– 1∣z+1∣2\frac{1}{|z + 1|^{2}}

C

1∣z+1∣2\frac{1}{|z + 1|^{2}}

D

2∣z+1∣2\frac{\sqrt{2}}{|z + 1|^{2}}

Answer

0

Explanation

Solution

Sol. |z| = 1 Ž zzˉ\bar{z} = 1

2Re (P) = p + pˉ\bar{p} = z–1z+1\frac{z–1}{z + 1} + zˉ–1zˉ+1\frac{\bar{z}–1}{\bar{z} + 1}

= z–1z+1\frac{z–1}{z + 1} + 1z–11z+1\frac{\frac{1}{z}–1}{\frac{1}{z} + 1}= z–1z+1\frac{z–1}{z + 1} + 1–z1+z\frac{1–z}{1 + z} = 0

Ž Re (p) = 0