- #1

- 21

- 0

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

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- Thread starter string_656
- Start date

- #1

- 21

- 0

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

Petek

Gold Member

- 364

- 9

Are you familiar with the log function for complex variables?

- #3

- 21

- 0

hmmm no... that sounds like that could help

- #4

- 21

- 0

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

- #5

Petek

Gold Member

- 364

- 9

Note: some books use

HTH

Petek

- #6

- 21

- 0

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

lurflurf

Homework Helper

- 2,440

- 138

- #8

lurflurf

Homework Helper

- 2,440

- 138

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

- #9

- 21

- 0

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

- 21

- 0

opps that didnt post right

wouldnt i go z = ln(i-1)/(0 - i) *(0+i)/(0+i)

wouldnt i go z = ln(i-1)/(0 - i) *(0+i)/(0+i)

- #11

lurflurf

Homework Helper

- 2,440

- 138

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

- 21

- 0

oh, ok thanks.

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