How does

-j = e^(-j*(pi/2))??

Calculator says it's true but I tried in WolframAlpha and it says an whole bunch of non-interesting (for me) stuff but that it is a true statement.

Also, how

??


Edited,

Ok, I got the second one. Still need to know about the 1st one!

 

There are many ways to show this.

Are you comfortable with

\(
e^{j \Theta} \; = \; \cos \( \Theta \) \; + \; j \sin \( \Theta \)
\)

 

The locus of all points of the form \(e^{-j\theta}\) in the complex plane, is the unit circle, for all values of theta in the range \([-\pi, ..., \pi]\)
The point (0, -j) is on the unit circle and can be expressed in the given form.

 

How does

-j = e^(-j*(pi/2))??

Calculator says it's true but I tried in WolframAlpha and it says an whole bunch of non-interesting (for me) stuff but that it is a true statement.

Also, how

??


Edited,

Ok, I got the second one. Still need to know about the 1st one!

A simple answer is that the real part is 0 and the imaginary part is -1.
This is true of ANY expression, that you can find the real and imaginary parts and sometimes you will get -1 for the imaginary part and 0 for the real part. More generally, you'll get something for the real part and something for the imaginary part.
For example:
e^(i*pi/4-pi/4);

the real part is:
e^(-pi/4)/sqrt(2)

and the imaginary part is:
e^(-pi/4)/sqrt(2)

so the result is:
i*e^(-pi/4)/sqrt(2)+e^(-pi/4)/sqrt(2)

or in floats approximately:
0.3223969419*i+0.3223969419

so the exponential form is the exponential form of a complex number.

 

There are many ways to show this.

Are you comfortable with

\(e^{j \Theta} \; = \; \cos \( \Theta \) \; + \; j \sin \( \Theta \)
\)

I am.

I already got it.
It's just a matter of summing the powers of the exponential, like:

\(e^{-j\cdot \frac{\pi }{2}}\cdot e^{j\cdot \frac{\pi }{4}} = e^{j\left ( \frac{\pi }{4}-\frac{\pi }{2} \right )}=e^{-j\frac{\pi }{4}}\)