There is an example in my textbook (Mathematical Methods for Physicists) that goes like this:

$$z=x+iy=r\cos\theta+ir\sin\theta=r(\cos\theta+i\sin\theta)=re^{i\theta}$$ It states that $re^{i\theta}$ is the polar form of a complex number, and $x+iy$ is the rectangular form of the complex number. Now I follow along with this just fine, but it made me confused about the other examples of the complex exponentials. Is something like $e^{{i\pi\over 2}}=i$ also in polar form? I think my confusion may come from there being an angle in the exponential, and I'm kind of rusty on polar coordinates (I've been working on that, though!).

  • 1
    $\begingroup$ Concerning $e^{{i\pi\over 2}}=i$: the LHS is in polar form, the RHS isn't. They're still equal, though. :-) $\endgroup$
    – user436658
    Oct 8, 2017 at 19:15
  • $\begingroup$ So does that mean that every time $e$ is raised to an angle multiplied by $i$ that is in polar form? $\endgroup$
    – matryoshka
    Oct 8, 2017 at 19:19
  • 1
    $\begingroup$ I think you have properly identified the source of your confusion. You can represent a complex number in rectangular coordinates with its real and imaginary parts, or in polar coordinates with radius and angle. The exponentiation does the right thing with the angle - that's Euler's formula. So polish up your polar coordinates and you'll be fine. $\endgroup$ Oct 8, 2017 at 19:22

4 Answers 4


For $z=x+iy\quad$ the norm is given by $|z|^2=z\bar z=(x+iy)(x-iy)=x^2+y^2$

Why is this important?

Because if we set $r=\sqrt{x^2+y^2}$

Then we arrive at $z=x+iy=\dfrac rr\times(x+iy)=r\left(\dfrac xr+i\dfrac yr\right)=r\ (a+ib)$

  • What can we say about $a=\dfrac xr$ and $b=\dfrac yr$ ?

Notice that $a^2+b^2=\dfrac{x^2}{r^2}+\dfrac{y^2}{r^2}=\dfrac{x^2+y^2}{r^2}=1$

Now $a^2+b^2=1$ is the equation of the circle of center $(0,0)$ and radius $1$ i.e. the unit circle.

And as you know, a point on the unit circle can be represented by $(\cos \theta,\sin \theta)$

Finally $a=\cos \theta$ and $b=\sin \theta$ for some $\theta$

And $z=x+iy=r\ (\cos \theta+i\sin\theta)=r\ e^{i\theta}$

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So given a non-zero point $(x,y)$ in rectangular form $z=x+iy$, it is always possible to find $(r,\theta)$ in polar form $z=r\ e^{i\theta}$.

When $z=0$, we cannot determine a unique $\theta$ since we can't divide by $r=0$, but $z=0=0\ e^{ia}$ with any $a$ is a valid polar form anyway.

So for $z=i=0+1\,i$ in rectangular form you have $r=\sqrt{0^2+1^2}=1$ and $a=0$ and $b=1$.

Now solving $\begin{cases}\cos\theta=a=0\\\sin\theta=b=1\end{cases}\implies \theta=\dfrac{\pi}2$ if we limit ourselves to $[0,2\pi[$.

And the polar form is $z=re^{i\theta}=e^{\frac{i\pi}2}$

  • $\begingroup$ edit 24/10: added a picture. $\endgroup$
    – zwim
    Oct 24, 2017 at 20:30

The polar form means that the complex number is known by its modulus ($r$) and its argument ($\theta$), rather than by its real and imaginary parts.

You can write equivalently

$$z=re^{i\theta}=r\angle\theta=r\text{ cis}(\theta).$$

In $e^{i\pi/2}$, you obviously have $r=1$ and $\theta=\pi/2$.


To answer your question concerning $e^{\frac{i\pi}{2}} = i$, neither side is in explicit polar form. To see this more clearly, use euler's formula

$e^{\frac{i\pi}{2}} = \text{cos}(\frac{\pi}{2}) + i\text{sin}(\frac{\pi}{2}) = 0 + i(1) = i$

EDIT: To clarify, the L.H.S (of your equation) is known as the exponential form of a complex number



Any complex number can be expressed in rectangular or polar form. So the number $i$ (in rectangular form : $0+1 i$) is the same as $e^{i\frac{\pi}{2}}$ ( in polar form with $r=1$ and $\theta =\frac{\pi}{2}$ ).

But note that is a one-one correspondence only if we limits $0 \le \theta < 2\pi$ because the exponential function in the complex field is periodic with period $2\pi i$.


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