# Simple complex number question

1. Jan 6, 2010

### PhyStan7

Sorry i can tell im being stupid and missing something here.

1. The problem statement, all variables and given/known data
if 7^(3+2i)=re^(i[theta]) find values of the real numbers r and [theta]

2. Relevant equations

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

3. The attempt at a solution

Ok you know that 7^(3+2i)=(7^3)(7^2i) but i dont 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 dont 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!

2. Jan 6, 2010

### tiny-tim

Hi PhyStan7!

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

Hint: 7i = (eln7)i

3. Jan 6, 2010

### PhyStan7

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

4. Jan 6, 2010

### tiny-tim

Yup, that's it!

5. Jan 6, 2010

### tomeatworld

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

?

6. Jan 7, 2010

### 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?

7. Jan 7, 2010

### tomeatworld

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

8. Jan 7, 2010

### Hurkyl

Staff Emeritus
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)

9. Jan 7, 2010

### tiny-tim

That's if a and b are real. ok then … if a/b is real, when is (a/b)i = 1 ?

10. Jan 7, 2010

### tomeatworld

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. Jan 7, 2010

### tiny-tim

uhh? That's b = ai.

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

12. Jan 7, 2010

### tomeatworld

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

13. Jan 7, 2010

### tiny-tim

Right!

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

14. Jan 7, 2010

### 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=$$e^{2n \pi}$$

15. Jan 7, 2010

### 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)

16. Jan 7, 2010

### tomeatworld

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