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Using Euler's formula

  • Thread starter ezperkins
  • Start date
  • #1
17
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"Use Euler's formula to evaluate the following and write your answer in rectangular form."
A. (2i)5
B. (1+i)-.5

I referred to my precal book and various websites and am still clueless. I started to work out A. but I'm not sure of anything. Here's what I did:

(2i)5 = 32i

On the imaginary/real plane, that forms a 90 degree angle.

[tex]\theta = \frac{\pi}{4}[/tex]

[tex]e^{i \theta } = cos \theta + isin \theta [/tex]

[tex] cos \frac{\pi}{4} = 0[/tex] & [tex]isin\frac{\pi}{4} = i [/tex]

[tex]e^ {\frac{i\pi}{4}} = i [/tex]

[tex]e^ {\frac{\pi}{4}} = ? [/tex] . . .

Whenever I don't know what I'm doing, I just mimic, and I feel like I'm mimicking incorrectly.
I would really like to know how to do this but can't figure it out on my own. Thanks in advance :)
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94
Well, firstly, a 90o angle isn't [itex]\pi /4[/itex], it's [itex]\pi /2[/itex] :tongue:

If [tex]e^{\frac{i\pi}{2}}=i[/tex]

then [tex]2e^{\frac{i\pi}{2}}=2i[/tex]

and [tex]\left(2e^{\frac{i\pi}{2}}\right)^5=(2i)^5[/tex]

Can you take it from here?
 
  • #3
17
0
haha thanks, I have a habit of making dumb mistakes like that.

I'm working it a few different ways but keep winding up with:

[tex] \cos \theta + i \sin \theta = i [/tex]

I've flown through all of the other problems on this stupid homework, but I've been working on this problem for about two hours and I still don't know what to do.
 
  • #4
Mentallic
Homework Helper
3,798
94
You're looking at the problem in entirely the wrong way.

You need to simplify:

[tex]\left(2e^{\frac{i\pi}{2}}\right)^5[/tex]

Do it like you would any other real number. What is [tex](ab^2)^3[/tex]?
 

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