0
$\begingroup$

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!).

$\endgroup$
3
  • 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 '17 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 '17 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 '17 at 19:22
1
$\begingroup$

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

enter image description here
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 $z=re^{i\theta}=e^{\frac{i\pi}2}$

$\endgroup$
1
  • $\begingroup$ edit 24/10: added a picture. $\endgroup$
    – zwim
    Oct 24 '17 at 20:30
1
$\begingroup$

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 $\theta=\pi/2$.

$\endgroup$
1
$\begingroup$

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

$z=|z|e^{i\theta}$

$\endgroup$
1
$\begingroup$

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 $2\pi i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.