Q21 of 66 Page 91

Solve the system of equations Re (z2) = 0, |z| = 2.

Given Re (z2) = 0, |z| = 2


Let z = x + iy.


Then


Given


⇒ x2 + y2 = 4 … (1)


Also, z2 = x2 + 2ixy – y2


= (x2 - y2) + 2ixy


Now, Re (z2) = 0


⇒ x2 – y2 = 0 … (2)


Solving (1) and (2), we get


⇒ x2 = y2 = 2


⇒ x = y = ±√2


∴ z = x + iy


= ±√2 ± i√2


Hence, we have four complex numbers.


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