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Question:

For all complex numbers z of the form 1+iα, α∈R, if z²=x+iy, then :

y²−x+2=0

y²+4x−0=0

y²−x+4=0

y²+4x+2=0

Solution:

It is given that z=1+iα
Hence, z²=(1+iα)²=(1−α²)+i(2α)=x+iy
Hence, x=1−α² and y=2α
Thus y²=4α²
Substituting in the equation of x gives us x=1−(y²/4)
4x=4−y²
y²+4x−4=0