If two complex numbers $a+bi$ and $c+di$ are equal then they are at the same point in the complex plane and $a=c$, $b=d$. The converse is also true, if $a=c$ and $b=d$ then $a+bi$ and $c+di$ are located at the same point in the complex plane and $a+bi=c+di$. Thus we may write:

$$a+bi=c+di\iff a=c, b=d$$

This can be used to split a single complex equation into two real equations that are then solved simultaneously. We illustrate the method by finding the square root of $i$.

Let

$$\sqrt{i}=a+bi$$

we square this equation to get

$$i=0+1i=(a+bi)^2=a^2+2abi+b^2i^2=(a^2\mathbin{-}b^2)+(2ab)i$$

equating the imaginary parts gives

$$2ab=1$$

equating the real parts gives

$$a^2\mathbin{-}b^2=0$$

or

$$a=\pm b$$

Using $a=b$ and $2ab=1$ gives $2b^2=1$ or $b=\pm\frac{1}{\sqrt{2}}=a$

Using $a=\mathbin{-}b$ and $2ab=1$ gives $b^2=\mathbin{-}\frac{1}{2}$ which is a dead end since $b$ is real.

Final solution is $a=b=\pm\frac{1}{\sqrt{2}}$

thus

$$\sqrt{i}=\pm\frac{1}{\sqrt{2}}\pm\frac{1}{\sqrt{2}}i$$

Fig. 3 shows the complex number $z=a+bi$ located at point $P$ and introduces two new variables $r$ and $\theta$

$0\leq r$ is the distance of $P$ from the origin $O$

$r$ is called the modulus or absolute value of the complex number z

$$r=\mod(z)=|z|$$

$0\leq\theta < 2\pi$ is the angle that $OP$ makes with the positive $x$ axis measuring in the anticlockwise direction. Alternatively it is the negative of the angle that $OP$ make with the $x$ axis measuring in the clockwise direction.

$\theta$ is called the angle, argument, amplitude or phase of $z$

$$\theta=\arg(z)$$

Note: The argument is unchanged if we add any integer multiple of $2\pi$ to the principal value $\theta$ ie. $\arg(z)=\theta+2\pi n$ where $n=0,\pm1,\pm2,\pm3,\cdots$

We can use the variables $r$ and $\theta$ to write a complex number $z$ in modulus-argument form thus

$$z=(r, \theta)$$

$a$, $b$, $r$, $\theta$ are related by the following transformation equations

$$r=|z| = |a+bi|=\sqrt{a^2+b^2}$$

$$\theta=\arg(z)=\arctan \left(\frac{b}{a}\right)$$

$$a=r\cos\theta$$

$$b=r\sin\theta$$

Let us now try to write $i$  in modulus argument form

$$i=0+1i$$

so we have $a=0$ and $b=1$

$$r=\sqrt{a^2+b^2}=\sqrt{0^2+1^2}=1$$

and

$$\theta=\arctan\left(\frac{b}{a}\right)=\arctan\left(\frac{1}{0}\right)=\arctan\infty$$

Thus we get $\theta=\frac{\pi}{2}$ measuring anticlockwise from the positive $x$ axis or $\theta=\mathbin{-}\frac{3\pi}{2}$ measuring clockwise from the positive $x$ axis. Now we can write $i$ in modulus-argument form as $(1,\frac{\pi}{2})$ or $(1,\mathbin{-}\frac{3\pi}{2})$.

The equations $a=r\cos\theta$ and $b=r\cos\theta$ give yet another way of writing complex numbers called the polar or trigonometric form

$$z=a+bi=r\cos\theta +ir\sin\theta = r(\cos\theta+i\sin\theta)$$

Finally we may use series for $\sin\theta$ and $\cos\theta$ to write

$$\cos\theta +i\sin\theta = \left(1\mathbin{-}\frac{\theta^2}{2!}+\frac{\theta^4}{4!}+\cdots\right) +i\left(\theta\mathbin{-}\frac{\theta^3}{3!}+\frac{\theta^5}{5!}\mathbin{-}\cdots\right)= 1+i\theta+\frac{{(i\theta)}^2}{2!}+\frac{{(i\theta)}^3}{3!}+\frac{{(i\theta)}^4}{4!}+\frac{{(i\theta)}^5}{5!}\cdots$$

Thus we arrive at Euler’s identity

$$\cos\theta +i\sin\theta = e^{i\theta}$$

So a complex number can be written as 

$$z=re^{i\theta}$$

This is called the Exponential or Euler form of a complex number and may also be written as 

$$z=|z|e^{i\arg(z)}$$

To write a number in exponential from all we need to know are its modulus $r$ and its phase $\theta$.

If the modulus $r=0$ then whatever the phase is we get $z=0$ and the point is at the origin of the complex plane.

If $r=1$ then $z=e^{i\theta}$ lies on the unit circle centered on the origin. There are infinitely such points but the following four stand out.

when $\theta=0$ we have $z=e^{i0}=\cos0+i\sin0=1+0i=1$

when $\theta=\frac{\pi}{2}$ we have $z=e^{i\frac{\pi}{2}}=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=0+1i=i$

when $\theta=\pi$ we have $z=e^{i\pi}=\cos\pi+i\sin\pi=\mathbin{-}1+0i=\mathbin{-}1$

when $\theta=\frac{3}{2}\pi$ we have $z=e^{i\frac{3}{2}\pi}=\cos\frac{3}{2}\pi+i\sin\frac{3}{2}\pi=0\mathbin{-}1i=\mathbin{-}i$

These points are shown in Fig. 4.