# Complex number problem

1. Apr 24, 2004

### galipop

Hi All,

I've been asked to determine the values of z that obey the equation

e^z = 1 + sqrt(3)i

I'm still not sure the concept of this question. Could someone point me in the right direction?

Thanks

2. Apr 24, 2004

### matt grime

z=x+iy, put it in, e^x is then the modulus, and is uniquely determined, y isn't, but you can work out all the possibilities.

3. Apr 24, 2004

### moshek

well you can use the folowing direction:

e^(a+bi)=(e^a)*( cos(b)+sin(b)i)

Now you need to find a and b according to the result you want.
please notice that there are infinity number of solution to that !

Best
Moshek

4. Apr 24, 2004

### galipop

Thanks for the info...

So based on the first 2 replies:

e^(x+iy) = 1 + sqrt(3)i
e^x.e^iy = e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i

therefore e^x = sqrt(1*1 + sqrt(3)*sqrt(3))

so e^x = 2. which is unique.

Now I just need to find y right?

so am I on the right track?

Thanks

5. Apr 25, 2004

### matt grime

Yes, you're on the right track, you need to find the arguments.

6. Apr 25, 2004

### galipop

ok so next ...

tan(y) = b/a
tan(y) = sqrt(3)/1
y= PI/3

you talked about infinite solutions, so then

y = PI/3 + (PI*n) where n is an integer.

so now is have

x = ln 2
and
y= PI/3 + (PI*n)

so now do I just subsitute x and y into this eqn?
e^x*(cos(y)+sin(y)i) = 1 + sqrt(3)i

7. Apr 25, 2004

### HallsofIvy

Staff Emeritus
And after all that work, you arrive at: The solution to e^z = 1 + sqrt(3)i

is z= log(1+ sqrt(3)) since log z= log|z|+ arg(z)+ 2&pi;n

That is, 1+ sqrt(3)i, in "polar form" is 2 e^(2&pi;i/3) so
z= log(2)+ 2&pi;/3+ 2&pi;n where n is any integer.