UPSC NDA 2 2016 Mathematics Solved Paper

Let z=x+iy
z2+|z|=0
⇒x2−y2+2ixy+√x2+y2=0
⇒x2−y2+√x2+y2=0......(1)
and 2xy=0⇒xy=0⇒x=0 or y=0
Now: For y=0 in eq. (1) we get:
x2+√x2=0
⇒x2+|x|=0
Clearly x2+|x| will always be greater than 0 for all x>0
Let x≤0
x2+|x|=0
⇒x2−x=0⇒x(x−1)=0
⇒x=0‌or‌(x−1)=0
⇒x=0  (∵x≤0)
∴z=0
For x=0 in eq.(1) we get,
−y2+√y2=0
−y2+|y|=0
If y>0,then
−y2+|y|=0
⇒−y2+y=0
⇒y=0,y=1
⇒y=1   (∵y>0)
∴z=i
If y<0,then
−y2+|y|=0
⇒−y2−y=0
⇒y=0,y=−1
⇒y=−1   (∵y<0)
∴z=−i
∴There are only 3 distinct solutions.