- The Exponential Form of a Complex Number

\(z=x+yi=r(\cos⁡θ+i \sin⁡θ)=re^{iθ}\)

LESSON 1
Express the following complex numbers in the form \(re^{iθ}\).
(i) \(z_1=1+i\)
(ii) \(\displaystyle z_1={1+i\over \sqrt{3}-i}\)
SOLUTION
(i) \(z_1=1+i\)
  • Determine the modulus of \(z_1\)
\(r=|z|=\sqrt{1^2+1^2}=\sqrt{2}\)
  • Determine the argument of \(z_2\)
\(\arg ⁡z=\tan^{-1}⁡(1)={π\over4}\)
  • Substitute into \(z=re^{iθ}\)
\(\displaystyle z=\sqrt{2} e^{{π\over4} i}\)
\(\displaystyle z_1={1+i\over \sqrt{3}-i}\)
  • Let \(z_2=1+i\) and \(z_3\sqrt{3}-i\). Therefore, \(\displaystyle z_1={z_2\over z_3}\).
  • Determine the moduli and arguments for \(z_2\) and \(z_3\)
\(r_2=|z_2 |=\sqrt{1^2+1^2}=\sqrt{2}\)
\(\displaystyle \arg⁡ z_2={π\over4}\)
\(r_3=|z_3|=\sqrt{(\sqrt{3})^2+(-1)^2}=2\)
\(\displaystyle \arg⁡ z_3=-\tan^{-1}\left({1\over \sqrt{3}}\right)=-{π\over6}\)
  • Use the results \(\displaystyle |z_1 |={|z_2|\over|z_3|}\) and \(\arg⁡ z_1=\arg⁡ z_2-\arg⁡ z_3\)
\(\displaystyle |z_1|={|z_2|\over|z_3|}={\sqrt{2}\over2}\)
\(\displaystyle \arg⁡ z_1=\arg⁡ z_2 - \arg⁡ z_3={π\over4}-\left(-{π\over6}\right)={5π\over12}\)
  • Substitute modulus and argument into \(re^{iθ}\)
\(\displaystyle z_1={\sqrt{2}\over2} e^{{5π\over12} i}\)
NB: If \(z=re^{iθ}\) then \(z^*=re^{-iθ}\)

LESSON 3
(i) Rewrite \((-1+i)^9\) in the form \(re^{iθ}\) where \(r=|z|\) and \(θ=\arg ⁡z\).
(ii) Hence, prove that \((-1+i)^9=16(-1+i)\).
SOLUTION
(i) Let \(z_1=-1+i\)
  • Rewrite \(z_1\) in the form \(re^{iθ}\)
\(r_1=\sqrt{(-1)^2+1^2}=\sqrt{2}\)
\(\displaystyle θ_1=\arg⁡ z_1=\pi-\tan^{-1}⁡\left({1\over1}\right)={3π\over4}\)
\(\displaystyle z_1=\sqrt{2} e^{{3π\over4} i}\)
  • We can now determine \((z_1)^9\)
\(\displaystyle (-1+i)^9=(\sqrt{2}e^{{3π\over4} i})^9\)
\(\displaystyle =(\sqrt{2})^9(e^{{3π\over4} i×9})\)
\(\displaystyle =16\sqrt{2} e^{{27π\over4} i}\)
NB: Since \(-π≤θ≤π\), we must convert \(\displaystyle {27π\over4}\) to an equivalent angle within the indicated range. \(\displaystyle {27π\over4}=6 {3\over4} π\) therefore we subtract \(6π\) (the whole number part)
\(\displaystyle (-1+i)^9=16\sqrt{2}e^{{3π\over4}}\)
\(\displaystyle (-1+i)^9=16\sqrt{2}e^{{3π\over4}}\)
\(\displaystyle =16\sqrt{2} \left(\cos⁡\left({3π\over4}\right)+i \sin⁡\left({3π\over4}\right)\right)\)
\(\displaystyle =16\sqrt{2} \left(-{1\over \sqrt{2}}+{1\over \sqrt{2}}i\right)\)
\(=16(-1+i)\)