Roots of Complex Numbers (DP IB Analysis & Approaches (AA): HL): Revision Note

Roots of complex numbers

How do I find the square roots of a complex number?

• A complex number $z$ has two square roots, $w$ and $-w$, that are also complex

  • To find them, let $w=a+b\mathrm{i}$ where $a,b\in \mathrm{ℝ}$

  • and form simultaneous equations from the relationship $w^2=z$

    • by expanding $w^2$

    • and equating the real and imaginary parts with $z$

• E.g. find the square roots of $3+4\mathrm{i}$

  • Let $a+b\mathrm{i}$ be one of the square roots of $3+4\mathrm{i}$

  • This means ${\left(a+b\mathrm{i}\right)}^2=3+4\mathrm{i}$

  • Expand the left-hand side

    • $\left(a+b\mathrm{i}\right)\left(a+b\mathrm{i}\right)=a^2+ab\mathrm{i}+ba\mathrm{i}+b^2{\mathrm{i}}^2=a^2+2ab\mathrm{i}-b^2$

    • so $a^2+2ab\mathrm{i}-b^2=3+4\mathrm{i}$

  • Equate the real parts on both sides

    • $a^2-b^2=3$

  • Equate the imaginary parts on both sides

    • $2ab=4$

  • Solve these two equations simultaneously

    • e.g. make $b$ the subject of $2ab=4$ to get $b=\frac{2}{a}$

    • Substitute $b=\frac{2}{a}$ into $a^2-b^2=3$ to get $a^2-{\left(\frac{2}{a}\right)}^2=3$

    • Rearrange to $a^4-3a^2-4=0$ which gives $a^2=4$ or $a^2=-1$ (but $a\in \mathrm{ℝ}$)

    • so $a=\pm 2$ which gives $b=\pm 1$

  • The two square roots are $2+\mathrm{i}$ and $-2-\mathrm{i}$

How do I use de Moivre’s theorem to find the n^th roots of a complex number?

• De Moivre's theorem to find the $n$^th roots of a complex number, $z=r\left(\cos \theta +\mathrm{isin}\theta \right)$, is

  • $\sqrt[n]{z}=z^{\frac{1}{n}}=r^{\frac{1}{n}}\left(\cos \left(\frac{\theta +2k\pi }{n}\right)+\mathrm{i}\sin \left(\frac{\theta +2k\pi }{n}\right)\right)$

    • Letting $k=0,1,2,...,n-1$ gives each $n$^th root

• In exponential (Euler's) form, this is

  • $\sqrt[n]{z}=z^{\frac{1}{n}}=r^{\frac{1}{n}}{\mathrm{e}}^{\frac{\theta +2k\pi }{n}}$

    • where $k=0,1,2,...,n-1$

• e.g. find the fourth roots of $16\left(\cos \frac{\pi }{2}+\mathrm{i}\sin \frac{\pi }{2}\right)$

  • $\sqrt[4]{z}=z^{\frac{1}{4}}={16}^{\frac{1}{4}}\left(\cos \left(\frac{\frac{\pi }{2}+2k\pi }{4}\right)+\mathrm{i}\sin \left(\frac{\frac{\pi }{2}+2k\pi }{4}\right)\right)$

    • $k=0$ gives $2\left(\cos \frac{\pi }{8}+\mathrm{i}\sin \frac{\pi }{8}\right)$

    • $k=1$ gives $2\left(\cos \frac{5\pi }{8}+\mathrm{i}\sin \frac{5\pi }{8}\right)$

    • $k=2$ gives $2\left(\cos \frac{9\pi }{8}+\mathrm{i}\sin \frac{9\pi }{8}\right)$

    • $k=3$ gives $2\left(\cos \frac{13\pi }{8}+\mathrm{i}\sin \frac{13\pi }{8}\right)$

How are complex roots related to polygons?

• If you plot the $n$ complex roots of of a complex number on an Argand diagram

  • they form a regular $n$-sided polygon

• e.g. in the example above

  • $2\left(\cos \frac{\pi }{8}+\mathrm{i}\sin \frac{\pi }{8}\right)$, $2\left(\cos \frac{5\pi }{8}+\mathrm{i}\sin \frac{5\pi }{8}\right)$, $2\left(\cos \frac{9\pi }{8}+\mathrm{i}\sin \frac{9\pi }{8}\right)$ and $2\left(\cos \frac{13\pi }{8}+\mathrm{i}\sin \frac{13\pi }{8}\right)$

    • form a square

    • whose vertices lies on a circle of radius $r=2$