Square Root of a Complex Number

Square Root of a Complex Number

Consider $z=x+iy$, any complex number, and suppose $\sqrt{x+iy}=a+ib$ for some real $a,b$ we want to find. Squaring both sides:

Equating real and imaginary parts on the two sides gives two simultaneous real equations:

Solving these two equations together (typically with the help of the auxiliary identity $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=x^2+y^2$, which gives $a^2+b^2$ directly) yields the values of $a$ and $b$ — and hence the square root of $x+iy$.

Worked Example 1: find the square root of $6+8i$.

Let $\sqrt{6+8i}=a+ib$. Squaring: $6+8i=(a^2-b^2)+(2ab)i$. Equating parts: $6=a^2-b^2\ \ldots(1)$ and $8=2ab\ \ldots(2)$. From (2), $a=\dfrac4b$; substituting into (1): $6=\dfrac{16}{b^2}-b^2$, i.e. $6b^2=16-b^4$, i.e. $b^4+6b^2-16=0$. Setting $m=b^2$: $m^2+6m-16=0$, which factors as $(m+8)(m-2)=0$, so $m=-8$ or $m=2$. Since $b$ is real, $b^2\neq-8$, so $b^2=2$, giving $b=\pm\sqrt2$. For $b=\sqrt2$: $a=\dfrac{4}{\sqrt2}=2\sqrt2$. For $b=-\sqrt2$: $a=-2\sqrt2$. So $\sqrt{6+8i}=2\sqrt2+\sqrt2\,i=\sqrt2(2+i)$, or its negative $-\sqrt2(2+i)$: $\sqrt{6+8i}=\pm\sqrt2(2+i)$.

Worked Example 2: find the square root of $3-4i$.

Let $\sqrt{3-4i}=a+ib$. Squaring: $3-4i=(a^2-b^2)+(2ab)i$, giving $a^2-b^2=3$ and $2ab=-4$. Using the auxiliary identity: $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=3^2+(-4)^2=9+16=25$, so $a^2+b^2=5$. Solving $a^2+b^2=5$ together with $a^2-b^2=3$: adding gives $2a^2=8\Rightarrow a^2=4\Rightarrow a=\pm2$. For $a=2$: $b=\dfrac{-4}{2(2)}=-1$. For $a=-2$: $b=\dfrac{-4}{2(-2)}=1$. So $\sqrt{3-4i}=2-i$ or $-2+i$. …