# Exponential functions, and complex numbers

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

Petek
Gold Member
Are you familiar with the log function for complex variables?

hmmm no... that sounds like that could help

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

Petek
Gold Member
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

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?

lurflurf
Homework Helper
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)]

lurflurf
Homework Helper
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?

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

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.

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

lurflurf
Homework Helper
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

oh, ok thanks.