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?



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.


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*}

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?

  • $\begingroup$ Thank you very much. Excellent and very clear explanation. $\endgroup$ – BTestQ Jan 20 '16 at 0:31
  • $\begingroup$ @BTestQ: You're welcome! :-) $\endgroup$ – Markus Scheuer Jan 20 '16 at 4:26
  • $\begingroup$ Could it be a mistake, that Im operator is not "idempotent", since, if I apply it to a, say, 5+i4, I should get number 4, and since 4 is real, next time I apply Im to 4, I get 0 ? If shown in Cartesian coordinate system, first application of Im will project to imaginary axis, and every next time, there will be no need to "move" dot and it will remain on the same spot on imaginary axis. This seems odd, since from first approach I could conclude one thing and from second approach something else. I am sure there is some flaw in my thinking, and would like to know where. $\endgroup$ – BTestQ Jan 20 '16 at 21:00
  • $\begingroup$ @BTestQ: Great, BtestQ! You're right of course! :-) My thinking was too template driven. I will correct it soon. What do you think about the $\Re$ operator? $\endgroup$ – Markus Scheuer Jan 20 '16 at 21:08
  • 1
    $\begingroup$ R operator is idempotent, no doubt about that :). $\endgroup$ – BTestQ Jan 20 '16 at 21:46

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