Exercise 1575Algebra › Complex numbersL1Added 3 Nov 2026

Square roots of complex numbers

#square roots#complex#system

Statement

Compute the square roots of $z=-8+6i$ .

Hint

Look for $w=a+bi$ with $w^2=z$ . Use $a^2-b^2=\operatorname{Re}(z)$ and $2ab=\operatorname{Im}(z)$ .

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System. $a^2-b^2=-8$ , $2ab=6$ , so $ab=3$ . Also $|w|^2=|z|=\sqrt{64+36}=10$ , so $a^2+b^2=10$ .

Solve. $a^2=(10-8)/2=1$ , $b^2=(10+8)/2=9$ . With $ab=3>0$ , $a,b$ have the same sign: $(a,b)=(1,3)$ or $(-1,-3)$ .

Conclusion. The square roots are $\pm(1+3i)$ .

Check. $(1+3i)^2=1+6i+9i^2=-8+6i$ . Correct.