Simple complex number question

PhyStan7
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Sorry i can tell I am being stupid and missing something here.

Homework Statement


if 7^(3+2i)=re^(i[theta]) find values of the real numbers r and [theta]


Homework Equations



er^(i[theta])=r(cos[theta]+isin[theta])


The Attempt at a Solution



Ok you know that 7^(3+2i)=(7^3)(7^2i) but i don't know what to do from here. I think there must be a simple formula or concept i am missing :s. There are similar questions aswel such as (1+i)^i which i also don't know how to solve. I know it must involve getting the i off the top but am unsure how to go about this. Advice would be appreciated, Thanks!
 
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Hi PhyStan7! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)

Hint: 7i = (eln7)i :wink:
 
Cheers tiny-tim!

So is this right...

343e(2i)ln7=re

and as re=r(cos(θ)+isin(θ))

θ=ln49 and r=343?

Or have i gone wrong somewhere?

Cheers
 
Yup, that's it! :smile:
 
Hi, just thought I'd jump in and ask if there was a proof for 7i=(eln(7))i

Cheers!

edit: the better question is probably, should it proven like:

7i = e
i ln(7) = iθ
ln(7) = θ

?
 
Hi tomeatworld! :wink:

A more fundamental question is, when does ai = bi ?

That's the same as (a/b)i = 1, so put a/b in polar form, and what do you get? :smile:
 
Surely if ai = bi then a = b. Converting a/b to polar form, surely just leaves a/b. Rather confused...
 
tomeatworld said:
Hi, just thought I'd jump in and ask if there was a proof for 7i=(eln(7))i
Actually, in a typical formulation,
7i=exp(ln(7) i)​
is the definition of what complex exponentiation means. (with exp defined by its power series)
 
tomeatworld said:
Converting a/b to polar form, surely just leaves a/b.

That's if a and b are real. ok then … if a/b is real, when is (a/b)i = 1 ? :smile:
 
  • #10
Well, quick logging and algebra gives me it will equal 1 when: b=ei ln(a). Meaning that when a>0 it works. Am I missing anything or got it completely wrong?
 
  • #11
uhh? That's b = ai. :confused:

Try this … what are all the real solutions for xi = 1 ?
 
  • #12
So i ln(x) = 0. then if the imaginary part is 0, i spose you should say x=e^{2n \pi}. Sound right?
 
  • #13
tomeatworld said:
So i ln(x) = 0. then if the imaginary part is 0, i spose you should say x=e^{2n \pi}. Sound right?

Right! :smile:

ok, so, for general complex numbers a and b, ai = bi when … ? :wink:
 
  • #14
When you generalise a complex number into a letter a, I can't understand how you deal with it :/ Unless it's just something like:

a=b=e^{2n \pi}
 
  • #15
Sorry, but you'll have to get used to it.

A general complex number usually is represented by a single letter (usually "z").

(Just like a vector being represented by a single letter a)
 
  • #16
we usually use z = a + bi and work from there. but with two it's killing me...
 

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