# Exponential functions, and complex numbers

1. Sep 10, 2009

### string_656

hey im doing some questions outta a txt book, i sorta understand complex numbers, like multiplying and dividing, ..

The question asks to rearrange for z,
e^(iz) = i - 1

im not sure what to do with the exponential function.

thanks for the help

2. Sep 10, 2009

### Petek

Are you familiar with the log function for complex variables?

3. Sep 10, 2009

### string_656

hmmm no... that sounds like that could help

4. Sep 10, 2009

### string_656

.. does any 1 no how i could solve this?

5. Sep 10, 2009

### Petek

You want to take the log of both sides of the equation. Look in your text book for more information.

Note: some books use ln instead of log.

HTH

Petek

6. Sep 10, 2009

### string_656

hmm ok so that gives...

z = ln(i - 1)/i
it cant be that easy?

what about... since e^(iz) = cos(z) + isin(z)

cos(z) + isin(z) = i - 1
Then how would i solve for Z?

7. Sep 10, 2009

### lurflurf

Think about where the line segment between i - 1 and 1 intersects the unit circle. Use trigonometry since exp(z)=exp(x+y*i)=exp(x)*[cos(y)+i*sin(y)]

8. Sep 10, 2009

### lurflurf

z = ln(i - 1)/i
is right, it depends on the form you want the answer in

solving this will work but z will be complex
cos(z) + isin(z) = i - 1

write i-1=sqrt(2)(i*sqrt(2)/2-sqrt(2)/2)
then solve
exp(x)=sqrt(2)
and
cos(y) + isin(y)=(i*sqrt(2)/2-sqrt(2)/2)
y real so
cos(y)=-sqrt(2)/2
sin(y)=sqrt(2)/2
tan(y)=-1

9. Sep 10, 2009

### string_656

hmm ok .. just a question, when its iz = ln|i-1|

wouldnt i go z = ln(i-1) * (0+i)
(0 - i) * (0+i)
because im just remembering when you divide complex numbers thats all.

10. Sep 10, 2009

### string_656

opps that didnt post right
wouldnt i go z = ln(i-1)/(0 - i) *(0+i)/(0+i)

11. Sep 10, 2009

### lurflurf

exp(iz)=i-1
take log
log(exp(iz))=log(i-1)
log cancels exp
iz=log(i-1)
divide by i
z=log(i-1)/i
that is the priciple value , there are others

12. Sep 11, 2009

### string_656

oh, ok thanks.