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Question: Let z be a complex number (not lying on X-axis of maximum modulus such that \(|z| > 1\). Then....

Let z be a complex number (not lying on X-axis of maximum modulus such that ∣z∣>1|z| > 1. Then.

A

∣z∣<1|z| < 1

B

zz

C

∣z2∣=∣z∣2|z^{2}| = |z|^{2}

D

None of these

Answer

zz

Explanation

Solution

Let =2iy(x+1)2+y2= \frac{2iy}{(x + 1)^{2} + y^{2}}.

Then (z−1z+1)\left( \frac{z - 1}{z + 1} \right)=1

⇒ ∣2z−1∣+∣3z−2∣|2z - 1| + |3z - 2|.

⇒ ∣2z−1∣|2z - 1|

⇒z=12z = \frac{1}{2}

Since =0+12=12,= 0 + \frac{1}{2} = \frac{1}{2}, is maximum, therefore ∣3z−2∣|3z - 2|

Differentiating (i) w.r.t.z=23z = \frac{2}{3}, we get

=13+0=13= \frac{1}{3} + 0 = \frac{1}{3}

Putting 13\frac{1}{3},we get ∣z∣=1⇒∣x+iy∣=1⇒x2+y2=1|z| = 1 \Rightarrow |x + iy| = 1 \Rightarrow x^{2} + y^{2} = 1⇒ ω=z−1z+1=(x−1)+iy(x+1)+iy×(x+1)−iy(x+1)−iy\omega = \frac{z - 1}{z + 1} = \frac{(x - 1) + iy}{(x + 1) + iy} \times \frac{(x + 1) - iy}{(x + 1) - iy}or =(x2+y2−1)(x+1)2+y2+2iy(x+1)2+y2=2iy(x+1)2+y2= \frac{(x^{2} + y^{2} - 1)}{(x + 1)^{2} + y^{2}} + \frac{2iy}{(x + 1)^{2} + y^{2}} = \frac{2iy}{(x + 1)^{2} + y^{2}}

z is purely imaginary or purely real.

((∵x2+y2=1)(\because x^{2} + y^{2} = 1) is not given)