2 The exponential form

Since $z=r\left(cos\theta +isin\theta \right)$ and since ${\text{e}}^{i\theta }=cos\theta +isin\theta$ we therefore obtain another way in which to denote a complex number: $z=r{\text{e}}^{i\theta }$ , called the exponential form .

Task!

Express $z=3{\text{e}}^{i\pi ∕6}$ in Cartesian form, correct to 2 d.p.

Use Key Point 9:

Answer

If $z=r{\text{e}}^{i\theta }$ and $w=t{\text{e}}^{i\varphi }$ then find expressions for

1. $z^{-1}$
2. $z^{\ast }$
3. $zw$

Solution

1. If $z=r{\text{e}}^{i\theta }$ then $z^{-1}=\frac{1}{r{\text{e}}^{i\theta }}=\frac{1}{r}{\text{e}}^{-i\theta }$ using the normal rules for indices.
2. Working in polar form: if $z=r{\text{e}}^{i\theta }=r\left(cos\theta +isin\theta \right)$ then

   $z^{\ast }=r\left(cos\theta -isin\theta \right)=r\left(cos\left(-\theta \right)+isin\left(-\theta \right)\right)=r{\text{e}}^{-i\theta }$

   since $cos\left(-\theta \right)=cos\theta$ and $sin\left(-\theta \right)=-sin\theta$ . In fact this reflects the general rule: to find the complex conjugate of any expression simply replace $i$ by $-i$ wherever it occurs in the expression.

3. $zw=\left(r{\text{e}}^{i\theta }\right)\left(t{\text{e}}^{i\varphi }\right)=rt{\text{e}}^{i\theta }{\text{e}}^{i\varphi }=rt{\text{e}}^{i\theta +i\varphi }=rt{\text{e}}^{i\left(\theta +\varphi \right)}$ which is again the result we are familiar with: when complex numbers are multiplied their moduli multiply and their arguments add.

We see that in some circumstances the exponential form is even more convenient than the polar form since we need not worry about cumbersome trigonometric relations.

Task!

Express the following complex numbers in exponential form:

1. $z=1-i$
2. $z=2+3i$
3. $z=-6$ .

Answer

Answer

Answer