# find all values of z satisfying...

• May 31st 2011, 09:03 PM
linalg123
find all values of z satisfying...
$exp[2z] + exp[z] + 1 = 0$

Really not sure how to start this one, i tried breaking it into it's real and imaginary components but didn't get far with that.

$exp[2x] \cdot(cos(2y)+isin(2y)) + exp[x]\cdot(cos(y)+isin(y)) +1 = 0 ??$

Is it something to do with taking the log?
If so i'm a little lost on how to do it
• May 31st 2011, 09:07 PM
mr fantastic
Quote:

Originally Posted by linalg123
$exp[2z] + exp[z] + 1 = 0$

Really not sure how to start this one, i tried breaking it into it's real and imaginary components but didn't get far with that.

$exp[2x] \cdot(cos(2y)+isin(2y)) + exp[x]\cdot(cos(y)+isin(y)) +1 = 0 ??$

Is it something to do with taking the log?
If so i'm a little lost on how to do it

Start by substituting w = e^z and solving for w.
• Jun 1st 2011, 05:37 PM
linalg123
$w^2 + w + 1 = 0$

$w = \frac{-1 \pm \sqrt{-3}}{2}$

$w = \frac{-1 \pm 3i}{2}$

$z = \ln{\frac{-1 \pm 3i}{2}}$

not sure how to continue?
• Jun 1st 2011, 06:10 PM
TheEmptySet
Quote:

Originally Posted by linalg123
$w^2 + w + 1 = 0$

$w = \frac{-1 \pm \sqrt{-3}}{2}$

$w = \frac{-1 \pm 3i}{2}$

$z = \ln{\frac{-1 \pm 3i}{2}}$

not sure how to continue?

Write the numbers in polar form

$\frac{-1+i\sqrt{3}}{2}=e^{i\frac{5\pi}{6}} \implies e^{z}=e^{i\frac{5\pi}{6}}$
• Jun 1st 2011, 06:45 PM
mr fantastic
Quote:

Originally Posted by linalg123
$w^2 + w + 1 = 0$

$w = \frac{-1 \pm \sqrt{-3}}{2}$

$w = \frac{-1 \pm 3i}{2}$

$z = \ln{\frac{-1 \pm 3i}{2}}$

not sure how to continue?

Surely you've been taught about the complex log function ....?

You should know that $\ln (z) = \ln |z| + i (\theta + 2n \pi)$ where $\theta$ is the principle argument of z and n is an integer. (The z here is not to be confused with the z in your question, by the way). Go back and review your class notes and textbook.
• Jun 1st 2011, 07:22 PM
Prove It
Quote:

Originally Posted by linalg123
$w^2 + w + 1 = 0$

$w = \frac{-1 \pm \sqrt{-3}}{2}$

$w = \frac{-1 \pm 3i}{2}$

$z = \ln{\frac{-1 \pm 3i}{2}}$

not sure how to continue?

There is no need to use logarithms, if you note that $\displaystyle w = e^{z}$ and that you can write each of the values of $\displaystyle w$ in exponential form.
• Jun 1st 2011, 07:43 PM
linalg123
when i try that i get a messy answer.

$|z|= \sqrt{(\frac{-1}{2})^2 + (\frac{3}{2})^2)} = \sqrt{\frac{10}{4}}$

$\theta = \arctan{\frac{\frac{3}{2}}{\frac{1}{2}}} = \arctan{3}$

did i do something wrong?
• Jun 1st 2011, 08:03 PM
Prove It
Yes, the imaginary part of $\displaystyle w$ is $\displaystyle \pm \frac{\sqrt{3}}{2}$, not $\displaystyle \pm \frac{3}{2}$.
• Jun 1st 2011, 08:18 PM
linalg123
well that was dumb, thanks.

so it's $z= i(2n\pi \pm \frac{2\pi}{3})$ ?