Does the equation z2+(−2+2i)z−(1+2i)=0 have a real solution?
The answer is “no”. Note that z2+(−2+2i)z−(1+2i)=0⇔z2+(−2+2i)z=1+2i⇔z2+(−2+2i)z+(−2+2i2)2=1+2i+(−2+2i2)2⇔(z+−2+2i2)2=1+2i+(−1+i)2⇔(z+(−1+i))2=1+2i−2i⇔(z+(−1+i))2=1 Since the square roots of 1 are 1 and −1, we must have z=1−(−1+i)=2−i or z=−1−(−1+i)=−i. In summary, the solutions are 2−i and −i and neither is a real number.