Given z=1+iα,z2=x+iy
⇒(1+iα)2=x+iy
⇒12+i2α2+2iα=x+iy
Using i2=−1,
⇒1−α2+2iα=x+iy
Equating the real and imaginary parts, we get
x=1−α2 and y=2α
⇒α2=1−x and y2=4α2
⇒y2=4(1−x)
⇒y2=4−4x
⇒y2+4x−4=0.
JEE Main 2014 — Mathematics Algebra
For all complex numbers z of the form 1+iα,α∈R, if z2=x+iy, then
Held on 19 Apr 2014 · Verified 6 Jul 2026.
y2−4x+4=0
y2+4x−4=0
y2−4x+2=0
y2+4x+2=0
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