Engineering Math - Complex Roots and Powers

AI Thread Summary
The discussion revolves around evaluating the expression i^3 + i in the form of x + iy. Participants clarify that i can be expressed in polar form as reiθ, where for i, r equals 1 and θ equals π/2. There is confusion regarding the use of r in the context of complex numbers, with some contributors confirming that r represents the magnitude of the complex number. The conversation also touches on the logarithmic properties of complex numbers, specifically how to express ln(i) and its implications for calculations involving powers of i. Overall, the thread emphasizes the importance of understanding polar coordinates in complex number evaluations.
Tonik
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Homework Statement


Evaluate the following in x+iy form.
i3+i

Homework Equations


i=rei\Theta

The Attempt at a Solution


i=rei\frac{pi}{2}
i3+i = i3*ii
(-i)(ei2\frac{pi}{2})
=-ie-\frac{pi}{2}
=-.2079i

I understand how it all works out except \frac{pi}{2}. I can't figure out how they got \frac{pi}{2} in the first place. Any ideas? Thanks in advance.
 
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My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?
 
If y= i^i, then ln(y)= i ln(i). Further, i has modulus r= 1 and argument \theta= \pi/2 so that i= e^{i(\pi/2+ 2k\pi)} for k any integer (e^{2k\pi}= 1 for all integer k so these are all different ways of writing i). Then ln(i)= i(\pi/2+ 2k\pi) and so ln(y)= -(\pi/2+ 2k\pi).
 
NascentOxygen said:
My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?

The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos\Theta+isin\Theta)=rei\Theta
As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.
 
NascentOxygen said:
My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?

HallsofIvy said:
If y= i^i, then ln(y)= i ln(i). Further, i has modulus r= 1 and argument \theta= \pi/2 so that i= e^{i(\pi/2+ 2k\pi)} for k any integer (e^{2k\pi}= 1 for all integer k so these are all different ways of writing i). Then ln(i)= i(\pi/2+ 2k\pi) and so ln(y)= -(\pi/2+ 2k\pi).

Thank you both for the explanations! You're my heroes.
 
Tonik said:
The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos\Theta+isin\Theta)=rei\Theta
As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.
I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.
 
NascentOxygen said:
I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.

Yeah, I think you're probably right. Thanks again! :)
 

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