Solve for z in the equation e^z=0

**Physics-is-Phun** · February 1st 2011, 06:40 PM

This is a homework problem for my Mathematical Physics course, where we're learning complex numbers and so on right now. I didn't see the subject over in the Physics site, and I wasn't quite sure which category this belonged in over here, so please move it if it if it's more appropriate somewhere else!

Anyway, on to the problem.

Either solve for z in the equation $e^z = 0$ , or show that this cannot be done.

I attempted to show that it could not be done by taking the natural logarithm of both sides:

$ln(e^z) = ln(0)$

$z = ln(0)$

I then concluded by saying that, since you cannot take the natural logarithm of 0, the problem is unsolvable, or z is undefined, or you cannot solve for z.

I wanted to make sure that my reasoning is correct, or if I am actually wrong and it can be done (though I don't want the solution handed to me, if I'm wrong- just point me in the right direction, please!).

Thank you in advance!

**dwsmith** · February 1st 2011, 06:43 PM

Originally Posted by Physics-is-Phun

This is a homework problem for my Mathematical Physics course, where we're learning complex numbers and so on right now. I didn't see the subject over in the Physics site, and I wasn't quite sure which category this belonged in over here, so please move it if it if it's more appropriate somewhere else!

Anyway, on to the problem.

Either solve for z in the equation $e^z = 0$ , or show that this cannot be done.

I attempted to show that it could not be done by taking the natural logarithm of both sides:

$ln(e^z) = ln(0)$

$z = ln(0)$

I then concluded by saying that, since you cannot take the natural logarithm of 0, the problem is unsolvable, or z is undefined, or you cannot solve for z.

I wanted to make sure that my reasoning is correct, or if I am actually wrong and it can be done (though I don't want the solution handed to me, if I'm wrong- just point me in the right direction, please!).

Thank you in advance!

$e^z=e^x(\cos(y)+i\sin(y))$

$e^x>0$

When will cos(y) + isin(y) = 0

**Physics-is-Phun** · February 1st 2011, 07:03 PM

Not as far as I can tell- so that would indicate that you cannot solve for z and have it be meaningful/possible (although your method probably showed it much more clearly than mine)?

**dwsmith** · February 1st 2011, 07:10 PM

Originally Posted by Physics-is-Phun

Not as far as I can tell- so that would indicate that you cannot solve for z and have it be meaningful/possible (although your method probably showed it much more clearly than mine)?

$e^x(\cos(y)+i\sin(y))=e^z\neq 0$

**Physics-is-Phun** · February 1st 2011, 07:20 PM

Thank you very much! That was fast!

**HallsofIvy** · February 2nd 2011, 08:11 PM

Having said that $e^x$ is not 0, you still need to show that $cos(y)+ i sin(y)$ is not 0. That is, there there is no real number, y, such that both cos(y) and sin(y) are equal to 0.

**Physics-is-Phun** · February 2nd 2011, 08:27 PM

Isn't this already known, though? I thought that:

$sin(y) = cos(y - pi/2)$

If that's true, then if I want to try to get them to be equal:

$y = y - \pi/2$

and then:

$0 = \pi/2$

Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

(almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)

(edit- thank you, dwsmith, for the tip!)

**dwsmith** · February 2nd 2011, 08:30 PM

Originally Posted by Physics-is-Phun

Isn't this already known, though? I thought that:

$sin(y) = cos(y - pi/2)$

If that's true, then if I want to try to get them to be equal:

$y = y - pi/2$

and then:

$0 = pi/2$

Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

(almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)

Pi = \pi

You could do $\displaystyle \cos(y)=0 \ \ \ y=\frac{\pi}{2}+k\pi \ \ k\in\mathbb{Z}$

$\displaystyle \sin(y)=0 \ \ \ \ y=k\pi \ \ \ k\in\mathbb{Z}$

**HallsofIvy** · February 4th 2011, 04:42 AM

Originally Posted by Physics-is-Phun

Isn't this already known, though? I thought that:

$sin(y) = cos(y - pi/2)$

If that's true, then if I want to try to get them to be equal:

$y = y - \pi/2$

and then:

$0 = \pi/2$

Since that's not true, they cannot both equal 0... right? Was that what you were thinking of?

(almost irrelevant- I don't know how to make the "pi" symbol in the post. Does anyone know how, or if it's possible?)

(edit- thank you, dwsmith, for the tip!)

But I had no way of knowing if it was "already known" by you.

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