Thread: Can this be writen any differently?

1. Can this be writen any differently?

$\displaystyle (e^z-1)(e^z-i)=0$

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

I need to find all the solutions for $\displaystyle e^z$

2. Re: Can this be writen any differently?

Do you mean you want something of the form $\displaystyle e^z=...$? Yes, you're dealing with an equation of the form $\displaystyle a\cdot b=0$ which implies that $\displaystyle a=0 \ \mbox{or} \ b=0$ are the solutions.

3. Re: Can this be writen any differently?

I need to find the solutions for $\displaystyle e^z$.
I know how to find the solutions when it's written as $\displaystyle e^z=...$ but I have a hard time figuring this one out.

4. Re: Can this be writen any differently?

I gave you a hint:

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

5. Re: Can this be writen any differently?

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 $\displaystyle e^z$ has to be either 1 or i, right? But I'm not sure how to use this information.

6. Re: Can this be writen any differently?

If I solve for $\displaystyle e^z=1$ then I get the solution:
$\displaystyle z=2p \pi i$

If I solve for $\displaystyle e^z=i$ then I get the solution:
$\displaystyle z=i(\pi /2 + 2p \pi)$

7. Re: Can this be writen any differently?

Originally Posted by MathIsOhSoHard
$\displaystyle (e^z-1)(e^z-i)=0$
I need to find all the solutions for $\displaystyle e^z$
There are many, many solutions.
$\displaystyle e^z=1\text{ or }e^z=i$.

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

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

8. Re: Can this be writen any differently?

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

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

Then $\displaystyle z=\frac{i(4k+1)\pi}{2}.$
I understand the $\displaystyle z=i(2k\pi),~k\in\mathbb{Z}.$ that's the conclusion I came to too.
But I'm not sure I understand your second solution.

My book says the facit is:
$\displaystyle z=2k\pi i$
or
$\displaystyle z=1+2k\pi i$

9. Re: Can this be writen any differently?

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

10. Re: Can this be writen any differently?

Originally Posted by Plato
Check the text again because:
$\displaystyle e^{1+2ki\pi}=e~.$
This is taken straight out of my text book word by word (it is scanned):

Is it incorrect?

11. Re: Can this be writen any differently?

Originally Posted by MathIsOhSoHard
This is taken straight out of my text book word by word (it is scanned):

Is it incorrect?
If you have copied it correctly, $\displaystyle e$ to that power is $\displaystyle e\ne 1$.

12. Re: Can this be writen any differently?

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

$\displaystyle z=2p \pi i$ for when $\displaystyle e^z=1$

and

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

Any of these values will make one of the brackets 0, and thus the entire equation will be zero, right?

13. Re: Can this be writen any differently?

That is correct.