Exponential functions, and complex numbers

[string_656]

hey I am doing some questions outta a txt book, i sort of 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]

Are you familiar with the log function for complex variables?

 

[string_656]

hmmm no... that sounds like that could help

 

[string_656]

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

 

[Petek]

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

Note: some books use ln instead of log.

HTH

Petek

 

[string_656]

hmm ok so that gives...

z = ln(i - 1)/i
it can't 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]

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]

hmm ok so that gives...

z = ln(i - 1)/i
it can't 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

 

[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 I am just remembering when you divide complex numbers that's all.

 

[string_656]

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

 

[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 principle value , there are others

 

[string_656]

oh, ok thanks.