Complex Analysis - Exponential Form

[Destroxia]

Homework Statement

Write the given numbers in the polar form $re^{i\theta}$.

$\frac {2i} {(3e^{4+i})}$

Homework Equations

$z = re^(i\theta)$

$\theta = Arg(z)$

$r = |z| = \sqrt { x^2 + y^2 }$

The Attempt at a Solution

I'm not really sure how to go about the exponential on the bottom of form $e^z$. I've read through my book now about 10 times, and don't see any info on it in any of the chapters before the problem.

If you asked me I would say the bottom was already in polar form, but that doesn't seem correct due to the form being of $e^{x+iy}$ instead of $e^{i\theta}$. I also know I can rewrite $e^{x+iy}$ as $e^x(cos(y)+isin(y))$, I believe. But I'm not sure that is going to help me.

The only thing I can really do at this point is find $Arg(z)$ for $2i$ which ends up coming out to $+ \frac \pi 2$. I'm not sure how to algebraically manipulate these equations to combine them to the form of $x + iy$ so I can convert find my $Arg(z)$ for $e^{4+i}$, and then I can combine the Args through the property $Arg(\frac {z_1} {z_2}) = Arg(z_1) - Arg(z_2)$, as well as find the total $|z|$, and rewrite in polar form $|z|e^{i\theta}$

 

[Mark44]

Homework Statement

Write the given numbers in the polar form $re^{i\theta}$.

$\frac {2i} {(3e^{4+i})}$

The above is equal to $\frac {2i} 3 e^{-4 - i}$
Can you continue from there?

Homework Equations

$z = re^(i\theta)$

$\theta = Arg(z)$

$r = |z| = \sqrt { x^2 + y^2 }$

The Attempt at a Solution

I'm not really sure how to go about the exponential on the bottom of form $e^z$. I've read through my book now about 10 times, and don't see any info on it in any of the chapters before the problem.

If you asked me I would say the bottom was already in polar form, but that doesn't seem correct due to the form being of $e^{x+iy}$ instead of $e^{i\theta}$. I also know I can rewrite $e^{x+iy}$ as $e^x(cos(y)+isin(y))$, I believe. But I'm not sure that is going to help me.

The only thing I can really do at this point is find $Arg(z)$ for $2i$ which ends up coming out to $+ \frac \pi 2$. I'm not sure how to algebraically manipulate these equations to combine them to the form of $x + iy$ so I can convert find my $Arg(z)$ for $e^{4+i}$, and then I can combine the Args through the property $Arg(\frac {z_1} {z_2}) = Arg(z_1) - Arg(z_2)$, as well as find the total $|z|$, and rewrite in polar form $|z|e^{i\theta}$

 

[Destroxia]

The above is equal to $\frac {2i} 3 e^{-4 - i}$
Can you continue from there?

I can see how you got what you posted above, but I still am not sure how to deal with the exponential.

 

[Mark44]

I can see how you got what you posted above, but I still am not sure how to deal with the exponential.

$e^{a + b} = e^a \cdot e^b$

 

[Destroxia]

$e^{a + b} = e^a \cdot e^b$

So there is no way to get both the $i$ terms together?

 

[Destroxia]

I've come up with the solution, in case anyone views this thread with the same question.

You have to convert the top to polar first, which will then allow you to combine the i terms.

$\frac {2i} {3e^{4+i}} = \frac {2e^{\frac \pi 2 i}} {3e^4e^i} = \frac {2} {3e^4} e^{i(\frac \pi 2 -1)}$

 

[Mark44]

I've come up with the solution, in case anyone views this thread with the same question.

You have to convert the top to polar first, which will then allow you to combine the i terms.

$\frac {2i} {3e^{4+i}} = \frac {2e^{\frac \pi 2 i}} {3e^4e^i} = \frac {2} {3e^4} e^{i(\frac \pi 2 -1)}$

No, you don't have to convert the numerator to polar first.
$\frac {2i} {3e^{4+i}} = \frac 2 3 i e^{-4 - i} = \frac 2 3 i e^{-4}e^{-i} = \frac{2e^{-4}}3 ie^{-i}$
$= \frac{2e^{-4}}3 i(\cos(1) - i\sin(1)) = \frac{2e^{-4}}3 (sin(1) + i\cos(1) = \frac{2e^{-4}}3 (\cos(\pi/2 - 1) + i\sin(\pi/2 - 1)) = \frac{2e^{-4}}3 e^{i(\pi/2 - 1)}$