Why does multiplication act like scaling and rotation of a vector in the complex plane? [duplicate]

I regularly use the geometric analogy of multiplication by a complex number to represent a scaling and rotation of a vector in the complex plane. For a very simple example, i would point up along the Y axis and multiplying it by i again would be a 90 degree rotation resulting in something pointing in the -X direction.

The thing is, I no longer recall why this is true. It's not obvious to me any longer why multiplication is in any way connected to rotation (scaling seems fairly obvious) and I was unable to explain the logic behind this useful trick to a friend who asked why it worked.

Could I get a very clear explanation of this geometric interpretation of multiplication by complex numbers? I feel like it had to do with Euler's identity and the polar form of complex numbers but this math is quite a few years behind me.

Solution 1:

Every complex number can be written in the form $r e^{i \theta}$. This follows from Euler's identity $$ e^{i \theta} = \cos(\theta) + i \sin(\theta). $$ For a given complex number $z$, you can see how to pick $\theta$ and $r$ so that $z = r e^{i \theta}$. (By the way, you can prove Euler's identity by plugging $i \theta$ into the Taylor series for $e^x$ -- it's one of the most fun calculations in math.)

So, if $z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$, then $$ z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}. $$ In other words, to multiply two complex numbers, we "add the angles and multiply the lengths".

Edit:

Here is an alternative answer that avoids using Euler's identity. Even without Euler's identity, it's clear that any complex number $z$ can be written in the form $$ z = r(\cos \theta + i \sin \theta). $$ So, suppose that $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$. Now multiply $z_1$ and $z_2$: $$ z_1 z_2 = r_1 r_2 (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i r_1 r_2(\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2). $$ If you remember the addition formulas from trigonometry, we recognize that they have miraculously appeared on the right hand side. So we have discovered that $$ z_1 z_2 = r_1 r_2( \cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)). $$ This shows, again, that when we multiply complex numbers we add the angles and multiply the lengths.

Comments: One might object that in this alternative proof, the appearance of the sum formulas for sine and cosine seems like a miracle. So I probably prefer the first proof that uses Euler's identity. However, one might object that Euler's identity is itself a miracle, because when we plug $i \theta$ into the Taylor series for $e^x$, we are shocked to find that sine and cosine pop out. (At least, I don't see why I shouldn't be shocked.) When things work out this nicely, we know we have stumbled upon something perfect and beautiful.

Solution 2:

$\newcommand{\I}{\mathrm i}$Every complex number $z = x + y\I$ can be represented by a matrix: $$ \mathbf M = \begin{bmatrix} x &-y\\y&x \end{bmatrix} $$ It is easily checked that these matrices follow all rules of complex numbers. In particular, $\det \mathbf M = |z|^2$. If $w = u + v\I$ is another complex number, we have $$ \begin{bmatrix} x &-y\\y&x \end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix} = \begin{bmatrix} ux - vy\\uy + vx\end{bmatrix} $$ which is precisely the product $zw$. This shows complex numbers can both be treated as a vector on the plane and a linear transformation.

Every such $\mathbf M$ can be factored into a rotation matrix $\mathbf R$ and a scaling matrix $\mathbf S$, with (The case of $z = 0$ is excluded) $$ \mathbf{R} = \frac{1}{|z|} \begin{bmatrix} x &-y\\y&x \end{bmatrix}, \qquad \mathbf{S} = \begin{bmatrix} |z|&0\\0&|z| \end{bmatrix} $$ Then $\mathbf M = \mathbf R\mathbf S$.

Hence every complex multiplication can be seen as applying a rotation and a scaling to a vector on the plane.