# How do you reduce $H\left(e^{j\frac{\pi}{2}}\right)$ further according to a textbook solution

I want to know how I could get from the first line to the second. I've been trying to figure it out for a while with no luck. Thank you in advance!

\begin{align} H\left(e^{j0.5\pi}\right) &= \frac{1 - e^{-j0.5\pi}}{1 - 0.25e^{-j\pi}}\\ &=1.13e^{j\frac{\pi}{4}} \end{align}

• You need to review your complex arithmetic. Like need. Because it'll come up later. It needs to be almost as automatic as calculating 3/4. – TimWescott Feb 4 at 1:23
• Do you have any reference material to help re-familiarize myself with it? – nyxlotus Feb 4 at 2:51
• Oh man, it's been so long. I can tell you what room I was in (since torn down, it was Portland State University, in the building that looked like an inside-out shower stall), but not which class. – TimWescott Feb 4 at 5:54
• A web search on complex arithmetic coughed up a bunch of hits. If you're a video person, I'd suggest the Khan acadamy one, just by their reputation. – TimWescott Feb 4 at 5:57
• I recently came back from a year and a half long internship and forgot almost everything. So things have become a game of catch-up. I'll do my best to review all the material, thank you! – nyxlotus Feb 4 at 16:32

## 2 Answers

I got misled a bit in my initial reasoning. So, back to it.

If we have Equation $$(1)$$ here below \begin{align} H\left(e^{j0.5\pi}\right) &= \frac{1 - e^{-j0.5\pi}}{1 - 0.25e^{-j\pi}}\\ &= \frac{1 - e^{-j\frac{\pi}{2}}}{1 - \frac 14e^{-j\pi}}\tag{1}\\ \end{align} And we know that $$e^{j\theta} = \cos(\theta) + j\sin(\theta)$$. With this we have that \begin{align} e^{-j\frac{\pi}{2}} &= -j\\ e^{-j\pi} &= -1\\ e^{j\frac{\pi}{4}} &= \frac{\sqrt2}{2} + j\frac{\sqrt2}{2}\\ \end{align} Equation $$(1)$$ then becomes \begin{align} H\left(e^{j\frac{\pi}{2}}\right) &= \frac 45\cdot \left(1 + j\right)\\ &=\frac 45\cdot \left(\frac{\sqrt{2}\sqrt{2}}{2} + j\frac{\sqrt{2}\sqrt{2}}{2}\right)\\ &= \underbrace{\frac 45\cdot \sqrt{2}}_{\approx 1.13}\underbrace{\left(\frac{\sqrt{2}}{2} + j\frac{\sqrt{2}}{2}\right)}_{e^{j\frac{\pi}{4}}} \equiv 1.13\cdot e^{j\frac{\pi}{4}}\\ \end{align}

• $e^{-j\pi}=1$${}$? – Matt L. Feb 4 at 12:25
• Well spotted, my bad, let me fix that. – Gilles Feb 4 at 12:26
• But then it turns out that the original solution is correct, i.e., the magnitude is indeed $1.13$. – Matt L. Feb 4 at 12:31
• Back online, thanks for the sharp eye on it. – Gilles Feb 4 at 15:01
• Hey, thanks Gilles. Your and ZR Han's answer is helping me understand this a bit more. My class is challenging and I'm mainly using the textbook to learn the material. – nyxlotus Feb 4 at 16:34

By plotting the unit circle you will find that $$1-e^{-j\pi/2} = \sqrt{2}e^{j\pi/4}$$.

• I'm sorry I'm still having some trouble understanding how you got there, do you mind showing me your complete steps? I'm truly sorry to ask this of you. – nyxlotus Feb 4 at 2:55
• @nyxlotus Complex numbers behave just like vectors and follow the parallelogram law of vector addition. Here's a reference for you What Are Complex Numbers? – ZR Han Feb 4 at 4:29
• Thank you I'll do my best to review this infromation – nyxlotus Feb 4 at 16:31