# Can this be writen any differently?

• Nov 28th 2011, 08:04 AM
MathIsOhSoHard
Can this be writen any differently?
$(e^z-1)(e^z-i)=0$

Can $e^z$ be isolated in any way?

I need to find all the solutions for $e^z$
• Nov 28th 2011, 08:07 AM
Siron
Re: Can this be writen any differently?
Do you mean you want something of the form $e^z=...$? Yes, you're dealing with an equation of the form $a\cdot b=0$ which implies that $a=0 \ \mbox{or} \ b=0$ are the solutions.
• Nov 28th 2011, 08:08 AM
MathIsOhSoHard
Re: Can this be writen any differently?
I need to find the solutions for $e^z$.
I know how to find the solutions when it's written as $e^z=...$ but I have a hard time figuring this one out.
• Nov 28th 2011, 08:11 AM
Siron
Re: Can this be writen any differently?
I gave you a hint:

Quote:

Originally Posted by Siron
you're dealing with an equation of the form $a\cdot b=0$ which implies that $a=0 \ \mbox{or} \ b=0$ are the solutions.

Can you do something with this?
• Nov 28th 2011, 08:14 AM
MathIsOhSoHard
Re: Can this be writen any differently?
Quote:

Originally Posted by Siron
I gave you a hint:

Can you do something with this?

Hmmmm...
Well one of the brackets must be zero...
So $e^z$ has to be either 1 or i, right? But I'm not sure how to use this information.
• Nov 28th 2011, 08:20 AM
MathIsOhSoHard
Re: Can this be writen any differently?
If I solve for $e^z=1$ then I get the solution:
$z=2p \pi i$

If I solve for $e^z=i$ then I get the solution:
$z=i(\pi /2 + 2p \pi)$
• Nov 28th 2011, 08:22 AM
Plato
Re: Can this be writen any differently?
Quote:

Originally Posted by MathIsOhSoHard
$(e^z-1)(e^z-i)=0$
I need to find all the solutions for $e^z$

There are many, many solutions.
$e^z=1\text{ or }e^z=i$.

In the first case $z=i(2k\pi),~k\in\mathbb{Z}.$

Then $z=\frac{i(4k+1)\pi}{2}.$
• Nov 28th 2011, 08:27 AM
MathIsOhSoHard
Re: Can this be writen any differently?
Quote:

Originally Posted by Plato
There are many, many solutions.
$e^z=1\text{ or }e^z=i$.

In the first case $z=i(2k\pi),~k\in\mathbb{Z}.$

Then $z=\frac{i(4k+1)\pi}{2}.$

I understand the $z=i(2k\pi),~k\in\mathbb{Z}.$ that's the conclusion I came to too.
But I'm not sure I understand your second solution.

My book says the facit is:
$z=2k\pi i$
or
$z=1+2k\pi i$
• Nov 28th 2011, 08:38 AM
Plato
Re: Can this be writen any differently?
Quote:

Originally Posted by MathIsOhSoHard
I understand the $z=i(2k\pi),~k\in\mathbb{Z}.$
My book says the facit is:
$z=2k\pi i$ or $z=1+2k\pi i$

Check the text again because:
$e^{1+2ki\pi}=e~.$
• Nov 28th 2011, 08:42 AM
MathIsOhSoHard
Re: Can this be writen any differently?
Quote:

Originally Posted by Plato
Check the text again because:
$e^{1+2ki\pi}=e~.$

This is taken straight out of my text book word by word (it is scanned):
http://img20.imageshack.us/img20/1263/solutionsn.jpg

Is it incorrect?
• Nov 28th 2011, 08:48 AM
Plato
Re: Can this be writen any differently?
Quote:

Originally Posted by MathIsOhSoHard
This is taken straight out of my text book word by word (it is scanned):
http://img20.imageshack.us/img20/1263/solutionsn.jpg
Is it incorrect?

If you have copied it correctly, $e$ to that power is $e\ne 1$.
• Nov 28th 2011, 08:58 AM
MathIsOhSoHard
Re: Can this be writen any differently?
Quote:

Originally Posted by Plato
If you have copied it correctly, $e$ to that power is $e\ne 1$.

$z=2p \pi i$ for when $e^z=1$
$z=i(\pi /2 + 2p \pi)$ for when $e^z=i$