# The complex exponential function

• Aug 22nd 2009, 11:36 AM
Jones
The complex exponential function
Hi,

I need to find all $z~~\text{such that}~~ e^{z} = 1+i$

So let $z =a+bi$ then $e^{a}*e^{bi}$

Now we can write this as $e^{a}*(cos(b) + i sin(b))$
Also i know that the modulus is $\sqrt{2}$
• Aug 22nd 2009, 12:02 PM
HallsofIvy
Quote:

Originally Posted by Jones
Hi,

I need to find all $z~~\text{such that}~~ e^{z} = 1+i$

So let $z =a+bi$ then $e^{a}*e^{bi}$

Now we can write this as [tex]e^{a}*(cos(b) + i sin(b))
Also i know that the modulus is $\sqrt{2}$

Well, that's all you need! If z= a+bi, then $e^z= e^{a+ bi}= e^acos(b)+ i e^asin(b))= 1+ i$ so you must have $e^a cos(b)= 1$ and $e^a sin(b)= 1$. Solve for a and b.
• Aug 22nd 2009, 12:04 PM
skeeter
Quote:

Originally Posted by Jones
Hi,

I need to find all $z~~\text{such that}~~ e^{z} = 1+i$

So let $z =a+bi$ then $e^{a}*e^{bi}$

Now we can write this as [tex]e^{a}*(cos(b) + i sin(b))
Also i know that the modulus is $\sqrt{2}$

$z = 1 + i
$

$z = e^a(\cos{b} + i\sin{b})$

equate the real and imaginary parts ...

$e^a \cos{b} = 1$

$e^a \sin{b} = 1$

$\tan(b) = 1$

$b = \frac{\pi}{4} + 2k\pi$

$e^a = \sqrt{2}$

$a = \frac{\ln{2}}{2}$
• Aug 22nd 2009, 12:32 PM
Plato
$z=\log(1+i)=\frac{\ln(2)}{2}+i\left(\frac{\pi}{4} +2n\pi\right).$
• Aug 22nd 2009, 12:56 PM
Jones
Thank you, i honestly thought it was much harder than this :)