Complex exponential expressions.

[lucasfish1]

Homework Statement

I just need some kind of explanation in layman's terms of what exactly is going on here. It seems as though I am missing some key element from trig. I am in a Signals class and the book lacks an explanation of the reduction used and ultimately why.

Homework Equations

I have attached a .pdf of the files.
View attachment Document1.pdf

The Attempt at a Solution

Ok, so in "b": I do not understand the simplification or reduction to the side, where 4pi is added. Also I think the same method occurs in the third answer of "d".

Also, is the reason there are 3 answers for "d" because of the cube root? If someone could just provide a short explanation as to why, that would be helpful.


Thank you so much in advance!

 

[Mark44]

Homework Statement

I just need some kind of explanation in layman's terms of what exactly is going on here. It seems as though I am missing some key element from trig. I am in a Signals class and the book lacks an explanation of the reduction used and ultimately why.

Homework Equations

I have attached a .pdf of the files.
View attachment 50347

The Attempt at a Solution

Ok, so in "b": I do not understand the simplification or reduction to the side, where 4pi is added. Also I think the same method occurs in the third answer of "d".

What they're doing in b is adding multiples of 2$\pi$ so as to get a positive angle.

From Euler's formula, e^ix = cos(x) + i*sin(x). This represents a complex number whose magnitude is 1, and that makes an angle of x radians measured counterclockwise from the positive Real axis. Note that mathematicians use i for the imaginary unit, and engineers use j.

Adding 2$\pi$ or multiples of 2$\pi$ to the exponent to get e^{i(x + 2$\pi$)} doesn't change anything except the angle. This complex number still has a magnitude of 1 - the only difference is that the angle is now x + 2$\pi$, which gets you to exactly the same point on the unit circle.

Also, is the reason there are 3 answers for "d" because of the cube root? If someone could just provide a short explanation as to why, that would be helpful.

In the same way that a number has two square roots, it will have three cube roots, four fourth roots, and so on. A real number, such as 8 has one real cube root (2) and two complex cube roots.

Thank you so much in advance!

 

[lucasfish1]

So, it is just commonplace to work within the unit circle? I mean for my calculator both arguments produce the same answer, I just appropriate the negative when using Euler's formula.

Thanks again.

Oh and by the way, I am on my way to be an EE so its "j" for me -here on out!

 

[Bacle2]

The complex exponential is periodic with (argument) period equal to 2Pi. So if you

go around n times, then e^z= e^z+2n∏ for n=1,2,... and

e^z+2n∏ is equivalent to going around a circle n times.