# Complex Analysis - Exponential Form

• Destroxia
In summary, the student attempted to solve for the exponential on the bottom of the equation, but was not sure how to do it. They found that the top was in polar form, and solved for the argumen t by combining the i terms.
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}##

RyanTAsher said:

## 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?
RyanTAsher said:

## 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 said:
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.

RyanTAsher said:
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##

Mark44 said:
##e^{a + b} = e^a \cdot e^b##

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

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)} ##

RyanTAsher said:
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)}##

## 1. What is the definition of the exponential form in complex analysis?

The exponential form in complex analysis is a way of representing a complex number in the form reix, where r is the magnitude of the number and x is the angle in radians. This form is also known as the polar form.

## 2. How is the exponential form used in complex analysis?

The exponential form is useful because it allows for easier calculations involving complex numbers, such as multiplication, division, and powers. It also provides a geometric interpretation of complex numbers, with the magnitude representing the distance from the origin and the angle representing the direction from the positive real axis.

## 3. Can complex numbers in exponential form be converted to rectangular form?

Yes, complex numbers in exponential form can be converted to rectangular form using the formula z = r(cos x + i sin x). This formula is known as Euler's formula and is a fundamental tool in complex analysis.

## 4. How does the concept of complex conjugates apply to exponential form?

In complex analysis, the complex conjugate of a number in exponential form is found by changing the sign of the angle. For example, if the number is reix, its complex conjugate would be re-ix. This property is useful in finding the real and imaginary parts of a complex number.

## 5. What are some applications of exponential form in real-world problems?

Exponential form is used in many areas of science and engineering, including electrical engineering, quantum mechanics, and signal processing. In these fields, complex numbers are used to represent quantities such as voltage, current, and phase shift. The exponential form makes it easier to analyze and solve complex problems involving these quantities.

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