Re: complex in the form a+ib

Quote:

Originally Posted by

**sigma1** hello all, i have the following question which i am trying to solve. express ln1+ e^i in the form a + ib.

what i have done.

ln both sides therefore

i = ln (a+ib)

i= ln a + ln ib

equating real and imaginary parts

ln a = 0

a= 1

and

i = ln ib

1= ln b

b= e

ans= 1+ie

is this correct? if not how would i appraoch the problem. thanks.

Some brackets would be nice. Are you asking to write $\displaystyle \displaystyle \ln{\left( 1 + e^i \right)} $ in the form $\displaystyle \displaystyle a + b\,i $?

Re: complex in the form a+ib

Hint :

$\displaystyle e^{ix}=\cos(x)+i\sin(x)$

Re: complex in the form a+ib

Quote:

Originally Posted by

**Prove It** Some brackets would be nice. Are you asking to write $\displaystyle \displaystyle \ln{\left( 1 + e^i \right)} $ in the form $\displaystyle \displaystyle a + b\,i $?

well the question does not have any brackets in it. am wondering if its an error. but how would you attempt to solve it without the brackets. that is ln1 + e^i .

Re: complex in the form a+ib

Well ln(1) = 0, so you're really just trying to simplify e^i, or if you like, e^(1i). Use the hint Princeps gave you.

Re: complex in the form a+ib

Quote:

Originally Posted by

**Prove It** Well ln(1) = 0, so you're really just trying to simplify e^i, or if you like, e^(1i). Use the hint Princeps gave you.

so the answer would be

cos1 + isin1 ?

Re: complex in the form a+ib

Re: complex in the form a+ib

Quote:

Originally Posted by

**Prove It** Correct

thanks alot.

Re: complex in the form a+ib

Quote:

Originally Posted by

**sigma1** thanks alot.

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