Simple complex number question

[PhyStan7]

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!

 

[tiny-tim]

Hi PhyStan7!

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

Hint: 7ⁱ = (e^ln7)ⁱ

 

[PhyStan7]

Cheers tiny-tim!

So is this right...

343e^(2i)ln7=re^iθ

and as re^iθ=r(cos(θ)+isin(θ))

θ=ln49 and r=343?

Or have i gone wrong somewhere?

Cheers

 

[tiny-tim]

Yup, that's it!

 

[tomeatworld]

Hi, just thought I'd jump in and ask if there was a proof for 7ⁱ=(e^ln(7))ⁱ

Cheers!

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

7ⁱ = e^iθ
i ln(7) = iθ
ln(7) = θ

?

 

[tiny-tim]

Hi tomeatworld!

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?

 

[tomeatworld]

Surely if aⁱ = bⁱ then a = b. Converting a/b to polar form, surely just leaves a/b. Rather confused...

 

[Hurkyl]

Hi, just thought I'd jump in and ask if there was a proof for 7ⁱ=(e^ln(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)

 

[tiny-tim]

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 ?

 

[tomeatworld]

Well, quick logging and algebra gives me it will equal 1 when: b=e^{i ln(a)}. Meaning that when a>0 it works. Am I missing anything or got it completely wrong?

 

[tiny-tim]

uhh? That's b = aⁱ.

Try this … what are all the real solutions for xⁱ = 1 ?

 

[tomeatworld]

So i ln(x) = 0. then if the imaginary part is 0, i spose you should say x=[tex]e^{2n \pi}[/tex]. Sound right?

 

[tiny-tim]

So i ln(x) = 0. then if the imaginary part is 0, i spose you should say x=[tex]e^{2n \pi}[/tex]. Sound right?

Right!

ok, so, for general complex numbers a and b, aⁱ = bⁱ when … ?

 

[tomeatworld]

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=[tex]e^{2n \pi}[/tex]

 

[tiny-tim]

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)

 

[tomeatworld]

we usually use z = a + bi and work from there. but with two it's killing me...