## - 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)
$$\displaystyle (-1+i)^9=16\sqrt{2}e^{{3π\over4}}$$
$$\displaystyle (-1+i)^9=16\sqrt{2}e^{{3π\over4}}$$
$$\displaystyle =16\sqrt{2} \left(\cos⁡\left({3π\over4}\right)+i \sin⁡\left({3π\over4}\right)\right)$$
$$\displaystyle =16\sqrt{2} \left(-{1\over \sqrt{2}}+{1\over \sqrt{2}}i\right)$$
$$=16(-1+i)$$