Engineering Math - Complex Roots and Powers

[Tonik]

Homework Statement

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

Homework Equations

i=re^i\Theta

The Attempt at a Solution

i=re^{i\frac{pi}{2}}
i³⁺ⁱ = i³*iⁱ
(-i)(e^{i2\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.

 

[NascentOxygen]

My editor won't allow greek letters, so in place of theta I'll use T

You know: e^iT=cosT + i.sinT

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

e^i.Pi/2= i

Q: How do you say in words: re^i.theta

I'm wondering what the r is part of?

 

[HallsofIvy]

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).

 

[Tonik]

My editor won't allow greek letters, so in place of theta I'll use T

You know: e^iT=cosT + i.sinT

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

e^i.Pi/2= i

Q: How do you say in words: re^i.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)=re^i\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.

 

[Tonik]

My editor won't allow greek letters, so in place of theta I'll use T

You know: e^iT=cosT + i.sinT

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

e^i.Pi/2= i

Q: How do you say in words: re^i.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).

Thank you both for the explanations! You're my heroes.

 

[NascentOxygen]

The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos\Theta+isin\Theta)=re^i\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.e^iT so that's when it's needed.

 

[Tonik]

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.e^iT so that's when it's needed.

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