Question

find the square root of i.Z​

Answer 1

Let z2 = (x + yi)2 = 8 – 6i

\ (x2 – y2) + 2xyi = 8 – 6i

Compare real parts and imaginary parts,

x2 – y2 = 8 (1)

2xy = -6 (2)

Now, consider the modulus: |z|2 = |z2|

\ x2 + y2 = Ö(82 + 62) = 10 (3)

Solving (1) and (3), we get x2 = 9 and y2 = 1

x = ±3 and y = ±1

From (2), x and y are of opposite sign,

\ (x = 3 and y = -1) or (x = -3 or y = 1)

\ z = ± (3 – i)

Answer 2

Answer:

$\sqrt{x {}^{2} + y {}^{2} }$

put z= X +iY