solve the system of equation

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

Let z = x + iy
Then z2 =  x2 + 2ixy - y2
And real(z2) = x2 - y2  = 0  (1)

And |z| = x2+y2
|z| = 2
Hence x2+y2 = 2  (2) 
Or x2 +  y2 = 4 (3)

Solving (1) and (3) , we get x = ±2 and y = ±2

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