Mathematics and Statistics · Ch 3 — Complex Numbers

Square Roots of a Complex Number

Every non-zero complex number has exactly two square roots, which are negatives of each other. They can be found algebraically without polar form.

The method

To find $\sqrt{a+bi}$ , set the root equal to $x+yi$ with $x,y$ real:

\sqrt{a+bi}=x+yi\ \Longrightarrow\ a+bi=(x+yi)^{2}=(x^{2}-y^{2})+2xy\,i.

Equating real and imaginary parts gives two equations:

x^{2}-y^{2}=a,\qquad 2xy=b.

A third, very convenient equation comes from moduli — since $|x+yi|^{2}=|{\sqrt{a+bi}}|^{2}=|a+bi|$ :

x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}.

Solving the system

Add and subtract the equations $x^{2}-y^{2}=a$ and $x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}$ :

x^{2}=\frac{\sqrt{a^{2}+b^{2}}+a}{2},\qquad y^{2}=\frac{\sqrt{a^{2}+b^{2}}-a}{2}.

Both right-hand sides are non-negative, so $x$ and $y$ are real. Finally the equation $2xy=b$ fixes the relative sign of $x$ and $y$ :

Note

The sign of $b$ pairs the roots …

Definition 1Square root of $a+bi$

Write $\sqrt{a+bi}=x+yi$ ; then $x^{2}-y^{2}=a$ , $2xy=b$ , and $x^{2}+y^{2}=\sqrt{a^{2}+b^{2}}$ . Solve for $x,y$ ; the sign of $b$ pairs them. The …

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