Solving |e^z| using complex numbers

• Oct 23rd 2010, 05:27 AM
SyNtHeSiS
Solving |e^z| using complex numbers
Solve $|e^z| = 2$ for z

Attempt:

$\sqrt{e^{2z}} = e^z$

Therefore:

$e^z = 2e^{0 + i2\pi k}$
$= e^{ln2 + i2\pi k}$

$z = ln2 +i2\pi k, k E Z$

but the correct answer was $z = ln2 + ik$
• Oct 23rd 2010, 05:35 AM
Plato
You need to notice that $\left| {e^z } \right| = \left| {e^{x + yi} } \right| = e^x \left| {\cos (y) + i\sin (y)} \right| = e^x$
• Oct 24th 2010, 02:51 AM
SyNtHeSiS
Quote:

Originally Posted by Plato
You need to notice that $\left| {e^z } \right| = \left| {e^{x + yi} } \right| = e^x \left| {\cos (y) + i\sin (y)} \right| = e^x$

But then how can you now solve for z like the orignal question asked to, since you dont have a z variable anymore?

Using what you told me I got:

$e^x$
$= 2$
$= e^{ln2}e^{0 + 2\pi k i}$

$x = ln2 + 2\pi k i, k E Z$

but my answer obviously isnt right.
• Oct 24th 2010, 03:58 AM
mr fantastic
Quote:

Originally Posted by SyNtHeSiS
But then how can you now solve for z like the orignal question asked to, since you dont have a z variable anymore?

Using what you told me I got:

$e^x$
$= 2$
$= e^{ln2}e^{0 + 2\pi k i}$

$x = ln2 + 2\pi k i, k E Z$

but my answer obviously isnt right.

You're expected to realise that x is real ....
• Oct 24th 2010, 05:07 AM
SyNtHeSiS
Oh I see, but why is it that you add $iy$ to the solution and not $2 \pi k i$?
• Oct 24th 2010, 05:35 AM
Plato
Quote:

Originally Posted by SyNtHeSiS
Oh I see, but why is it that you add $iy$ to the solution and not $2 \pi k i$?

No, see post #2.
The answer is $z=\ln(2)+yi$ for any $y\in \mathbb{R}.$
• Oct 24th 2010, 06:00 AM
SyNtHeSiS
Quote:

Originally Posted by Plato
No, see post #2.
The answer is $z=\ln(2)+yi$ for any $y\in \mathbb{R}.$

I understand what you did but can you just tell me why this method is wrong:

$2 = 2(cos0 + isin0)$
$= 2e^0$

I read somewhere in my notes that you always add $2\pi k i$to $\theta$ and in this case my $\theta$ is 0, so then I did the following:
$
= 2e^{0 + 2\pi k i}$

$e^x = e^{ln2}e^{2\pi k i}$

$x = ln2 + 2\pi k i, k E R$

I can see that k must be R since x is a real number.

But I just want to know why you dont add $2\pi ki$ here
• Oct 24th 2010, 06:10 AM
Plato
Quote:

Originally Posted by SyNtHeSiS
I read somewhere in my notes that you always add $2\pi k i$to $\theta$ and in this case my $\theta$ is 0, so then I did the following: $= 2e^{0 + 2\pi k i}$
$e^x = e^{ln2}e^{2\pi k i}$
$x = ln2 + 2\pi k i, k E R$
I can see that k must be R since x is a real number.
But I just want to know why you dont add $2\pi ki$ here

I don't know what your notes say, of course.
But you are missing the whole idea here.

$\left|e^{\ln(2)+yi}\right|=\left|e^{\ln(2)}\right| \left|\cos(y)+i\sin(y)\right|=2$ for all $y\in \mathbb{R}$.