Complex numbers: convert the exponential to polar form

[ac7597]

Summary:: Hello, my question asks if the complex exponential equation 4e^(-j) is equal to 4 ∠-180°. I tried to use polar/rectangular conversions: a+bj=c∠θ with c=(√a^2 +b^2) and θ=tan^(-1)[b/a]

4e^(-j)=4 ∠-180°
c=4, 4=(√a^2 +b^2)
solving for a : a=(√16-b^2)
θ=tan^(-1)[b/a]= -1
b/(√16-b^2)= tan(-1)
b=(√16-b^2)* tan(-1)
b=(√16-b^2)* -1.557
b^2=(16-b^2)(-1.557)^2
solving for b: b=3.365
4=(√a^2 +3.365^2) thus a=2.161
θ=tan^(-1)[-3.365/2.161]= -57.3°
Thus it is false

[Moderator's note: Moved from a technical forum and thus no template.]

 

[PeroK]

Summary:: Hello, my question asks if the complex exponential equation 4e^(-j) is equal to 4 ∠-180°.

What does this mean?

 

[etotheipi]

Summary:: Hello, my question asks if the complex exponential equation 4e^(-j) is equal to 4 ∠-180°.

Well one has an argument of $-1$, and the other an argument of $-\pi$...

 

[ac7597]

Yes, the question is rather 4e^(-1*j) = 4 ∠-180° ?

 

[PeroK]

Yes, the question is rather 4e^(-1*j) = 4 ∠-180° ?

Is $j = \sqrt{-1}$?

 

[etotheipi]

Yes, the question is rather 4e^(-1*j) = 4 ∠-180° ?

And if two complex numbers have different arguments, which aren't the same modulo $2\pi$, then are they equal?

Is $j = \sqrt{-1}$?

I believe it's an engineering notation

 

[ac7597]

Yes j=√-1

 

[berkeman]

Is $j = \sqrt{-1}$?

Yeah, it's electrical engineering notation. In EE "i" is used for current, so traditionally we use "j" for $\sqrt{-1}$

 

[PeroK]

Yes j=√-1

So, effectively, the question asks whether 180 degrees equals 1 radian?

 

[berkeman]

Yes, the question is rather 4e^(-1*j) = 4 ∠-180° ?

Are you familiar with the unit circle representation of a complex number...?

https://www.sciencedirect.com/topics/engineering/imaginary-axis

https://ars.els-cdn.com/content/image/3-s2.0-B9780750650489500052-f03-07-9780750650489.gif