⇒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.