Exponential Form of Complex Numbers
Euler's relation collapses the whole modulus–argument machinery into a single exponential — a third, even more compact way of writing a complex number.
From power series to Euler's relation
You already know that any complex number can be written $z=r(\cos\theta+i\sin\theta)$, where $r=|z|$ and $\theta=\arg z$. This lesson builds a third, even more compact way of writing $z$.
The Maclaurin series for $\cos\theta$, $\sin\theta$ and $e^x$ (which you'll meet properly in Chapter 2) are:
$$\cos\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots$$
$$\sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots$$
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
Substituting $x=i\theta$ into the series for $e^x$ and using $i^2=-1,\,i^3=-i,\,i^4=1$ to simplify each power:
$$e^{i\theta}=1+i\theta-\frac{\theta^2}{2!}-\frac{i\theta^3}{3!}+\frac{\theta^4}{4!}+\cdots$$
$$=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+i\left(\theta-\frac{\theta^3}{3!}+\cdots\right)$$
The bracketed real part is exactly the series for $\cos\theta$, and the bracketed imaginary part is exactly the series for $\sin\theta$. This gives Euler's relation:
$$e^{i\theta}=\cos\theta+i\sin\theta$$
Combining Euler's relation with the modulus-argument form gives the exponential form of a complex number:
$$z=re^{i\theta},$$
$$\text{where } r=|z|>0 \text{ and } \theta=\arg z \text{ satisfies } -\pi<\theta\leq\pi$$
As with the principal argument, $\theta$ is always chosen in this range unless a question says otherwise.
For $z$: comparing directly with $r(\cos\theta+i\sin\theta)$ gives $r=3$ and $\theta=\dfrac{2\pi}5$, so $z=3e^{\frac{2\pi}{5}i}$.
For $w$: first find $r=|w|=\sqrt{6^2+(-6)^2}=6\sqrt2$. The point $6-6i$ lies in the fourth quadrant, so $\theta=-\arctan\left(\dfrac66\right)=-\dfrac\pi4$. Therefore $w=6\sqrt2\,e^{-\frac\pi4 i}$.
Using Euler's relation, $z=4\left(\cos\dfrac{5\pi}6+i\sin\dfrac{5\pi}6\right)=4\left(-\dfrac{\sqrt3}2+\dfrac12 i\right)=-2\sqrt3+2i$.
If an angle you compute falls outside $-\pi<\theta\leq\pi$, add or subtract multiples of $2\pi$ until it lands back inside that range — exactly as you did with modulus-argument form. Since $\cos$ and $\sin$ are $2\pi$-periodic, this never changes the value of $e^{i\theta}$ itself.
Exam-style questions
Question 1 5 marks
Express $z=-4+4\sqrt3\,i$ in the form $re^{i\theta}$, where $-\pi<\theta\leq\pi$, giving $\theta$ as an exact multiple of $\pi$.
$r=\sqrt{(-4)^2+(4\sqrt3)^2}=\sqrt{16+48}=\sqrt{64}=8$. (M1, A1)
The point lies in the second quadrant. The reference angle is $\arctan\left(\dfrac{4\sqrt3}4\right)=\arctan(\sqrt3)=\dfrac\pi3$, so $\theta=\pi-\dfrac\pi3=\dfrac{2\pi}3$. (M1, A1)
$z=8e^{\frac{2\pi}{3}i}$. (A1)
Question 2 5 marks
Express $z=-3-3i$ in the form $re^{i\theta}$, where $-\pi<\theta\leq\pi$, giving $\theta$ as an exact multiple of $\pi$.
$r=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt2$. (M1, A1)
The point lies in the third quadrant. The reference angle is $\arctan\left(\dfrac33\right)=\dfrac\pi4$, so $\theta=-\left(\pi-\dfrac\pi4\right)=-\dfrac{3\pi}4$. (M1, A1)
$z=3\sqrt2\,e^{-\frac{3\pi}{4}i}$. (A1)