If a 0 and z = ((1+i)2/a-i) has magnitude √(2/5) then barz is equal to :

Question

If a > 0 and z = ((1+i)2/a-i) has magnitude √(2/5) then barz is equal to : (A)  - (3/5) - (1/5) i  (B)  - (1/5) + (3/5) i  (C)  - (1/5) - (3/5) i  (D)  (1/5) - (3/5) i

Tardigrade Admin · Accepted Answer

Given a > 0 z= ((1+i)2/a-i) = (2i(a+i)/a2+1) Also |z| = √(2/5) ⇒ (2/√a2+1) = √(2/5) ⇒ a=3 So barz = (-2i(3-i)/10) = (-1-3i/5)

Q. If $a > 0$ and $z = \frac{( 1 + i ) ^{2}}{a - i}$ has magnitude $\frac{2}{5}$ then $\overset{z}{ˉ}$ is equal to :

$- \frac{3}{5} - \frac{1}{5} i$

$- \frac{1}{5} + \frac{3}{5} i$

$- \frac{1}{5} - \frac{3}{5} i$

$\frac{1}{5} - \frac{3}{5} i$

Given $a > 0$
$z = \frac{( 1 + i ) ^{2}}{a - i} = \frac{2 i ( a + i )}{a ^{2} + 1}$
Also $∣ z ∣ = \frac{2}{5} \Rightarrow \frac{2}{a ^{2} + 1} = \frac{2}{5} \Rightarrow a = 3$
So $\overset{z}{ˉ} = \frac{- 2 i ( 3 - i )}{10} = \frac{- 1 - 3 i}{5}$

Q. If a>0 and z=a−i(1+i)2​ has magnitude 52​​ then zˉ is equal to :

−53​−51​i

−51​+53​i

−51​−53​i

51​−53​i