Engineering Math - Complex Roots and Powers

  • Thread starter Tonik
  • Start date
  • #1
13
0

Homework Statement


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

Homework Equations


i=rei[itex]\Theta[/itex]

The Attempt at a Solution


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

I understand how it all works out except [itex]\frac{pi}{2}[/itex]. I can't figure out how they got [itex]\frac{pi}{2}[/itex] in the first place. Any ideas? Thanks in advance.
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
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?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
964
If [itex]y= i^i[/itex], then [itex]ln(y)= i ln(i)[/itex]. Further, i has modulus r= 1 and argument [itex]\theta= \pi/2[/itex] so that [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for k any integer ([itex]e^{2k\pi}= 1[/itex] for all integer k so these are all different ways of writing i). Then [itex]ln(i)= i(\pi/2+ 2k\pi)[/itex] and so [itex]ln(y)= -(\pi/2+ 2k\pi)[/itex].
 
  • #4
13
0
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[itex]\Theta[/itex]+isin[itex]\Theta[/itex])=rei[itex]\Theta[/itex]
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.
 
  • #5
13
0
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 [itex]y= i^i[/itex], then [itex]ln(y)= i ln(i)[/itex]. Further, i has modulus r= 1 and argument [itex]\theta= \pi/2[/itex] so that [itex]i= e^{i(\pi/2+ 2k\pi)}[/itex] for k any integer ([itex]e^{2k\pi}= 1[/itex] for all integer k so these are all different ways of writing i). Then [itex]ln(i)= i(\pi/2+ 2k\pi)[/itex] and so [itex]ln(y)= -(\pi/2+ 2k\pi)[/itex].

Thank you both for the explanations! You're my heroes.
 
  • #6
NascentOxygen
Staff Emeritus
Science Advisor
9,244
1,072
The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos[itex]\Theta[/itex]+isin[itex]\Theta[/itex])=rei[itex]\Theta[/itex]
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.
 
  • #7
13
0
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! :)
 

Related Threads on Engineering Math - Complex Roots and Powers

  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
9
Views
2K
Replies
4
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
818
  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
2
Views
804
  • Last Post
Replies
4
Views
809
Replies
6
Views
7K
Top