Real and imaginary part functions

Question

How can real and imaginary part of $e^{ix}$ be used in complex tasks? I am aware that sine is expressed as imaginary part and cosine as real part, but I am confused when there is some sort of operation that is applied on these, how should I proceed? I suppose this comes from not knowing properties of this, should I say "operator" ($\text{Re},\text{Im}$).

Here is an example of my confusion: I have sequence $x_n = \sin^3(n\alpha)$, where $n$ is integer number, and $\alpha$ is some given angle. Using $\text{Im}$ function, that is same as $x_n = \text{Im}\{e^{in\alpha}\}^3$

With known system function $H(z)$, I should be able to find system response by plugging known $z$ and multiplying with $z^n$ but trouble is that I don't know what is my $z$ now. Is it $e^{i\alpha}$, where should I put that number 3?

I was thinking what if there is no number 3, let's say I just had to find response to sine, how should I use this $\text{Im}$ function (is this even right term?) when calculation $H(e^{i\alpha})$, if $H(z)$ looks something like $$H(z)=\frac{z(z+1)}{(z+2)(z+3)(z+4)}$$

Should I apply $\text{Im}$ function in front of fraction, or apply to every $z$ occurrence, or something third?

Answer 1

Operator:

You can consider both the Imaginary part \begin{align*} &\Im:\mathbb{C}->\mathbb{R} &\Im(z)=\Im(x+iy)=y\\ \end{align*} as operator as well as the Real part \begin{align*} &\Re:\mathbb{C}->\mathbb{R} &\Re(z)=\Re(x+iy)=x \end{align*}

In fact, the latter is a projection operator which orthogonally projects a complex number $x+iy$ to the $x$-axis. It has the property that repeated application of it gives the same result. \begin{align*} \Re(\Re(z))=\Re(\Re(x+iy))=\Re(x)=x \end{align*}

The $\Im$ operator does not share this property.

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Conjugation:

It's convenient to consider the easy to remember rules of complex conjugation

\begin{align*} \overline{z_1+z_2}&=\overline{z_1}+\overline{z_2}\\ \overline{z_1-z_2}&=\overline{z_1}-\overline{z_2}\\ \overline{z_1\cdot z_2}&=\overline{z_1}\cdot\overline{z_2}\\ \overline{\left(\frac{z_1}{z_2}\right)}&= \frac{\overline{z_1}}{\overline{z_2}}\\ \end{align*}

together with

\begin{align*} \Re(z)&=\frac{1}{2}\left(z+\overline{z}\right)\\ \Im(z)&=\frac{1}{2i}\left(z-\overline{z}\right)\\ \end{align*}

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Complicated situations: If we consider $H(z)$ in OPs example we obtain \begin{align*} \Im\left(H(z)\right)&=\frac{1}{2i}\left(H(z)+\overline{H(z)}\right) \end{align*} and by multiple application of the rules for complex conjugation \begin{align*} \overline{H(z)}&=\overline{\frac{z(z+1)}{(z+2)(z+3)(z+4)}}\\ &=\frac{\overline{z}(\overline{z}+1)}{(\overline{z}+2)(\overline{z}+3)(\overline{z}+4)} \end{align*}

Can you continue?