#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
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!

Last edited:

#### WBahn

Joined Mar 31, 2012
26,398
There are many ways to show this.

Are you comfortable with

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

#### Papabravo

Joined Feb 24, 2006
16,988
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.

#### MrAl

Joined Jun 17, 2014
8,501
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.

#### PsySc0rpi0n

Joined Mar 4, 2014
1,555
There are many ways to show this.

Are you comfortable with

$$e^{j \Theta} \; = \; \cos \( \Theta$$ \; + \; j \sin $$\Theta$$
\)
I am.

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