Finding the Polar Form of a Complex Number Using Euler's Relation

[aaj92]

Homework Statement

Using Euler's relation, prove that any complex number z=x+yi can be written in the form z= re$^{i\theta}$ where r and $\theta$ are real. Describe the significance of r and $\theta$ with reference to the complex plane.

b) Write z= 3+4i in the form z = re$^{i\theta}$
(pretty sure I can get this one if I can get help on the proof.

Homework Equations

e$^{i\theta}$= cos$\theta$+isin$\theta$

The Attempt at a Solution

I tried to prove it, got what it wanted me to get but I feel like I did it wrong because I don't know how to go about doing part b. there's also a part c but I didn't feel the need to put it up here because if someone can just explain to me the proof for these equations I think I should be able to get parts b and c

 

[genericusrnme]

How do cartesian coordinates relate to polar coordinates?

 

[aaj92]

x = rcos$\theta$
y = rsin$\theta$

...is that all you have to do?

so that makes sense, but I guess I was wrong about knowing how to do part b then... I don't know how to find r and $\theta$ given z = 3+4i

 

[genericusrnme]

yes
So you have your two equations
$x=r\ Cos( \theta)$
$y=r\ Sin( \theta)$

How would you find r in terms of x and y?

 

[aaj92]

r = $\frac{x}{cos\theta}$

r = i$\frac{y}{sin\theta}$ ??

 

[genericusrnme]

r = $\frac{x}{cos\theta}$

r = i$\frac{y}{sin\theta}$


??

nono, r in terms of x and y does not contain any mention of $\theta$

Make use of the fact that $Cos( \theta)^2 + Sin( \theta)^2 = 1$.
You should end up with pythagoras' theorem.

To find $\theta$, you can make use of $\frac{ Sin(\theta)}{Cos( \theta)} = Tan(\theta )$

 

[aaj92]

ok well I'm lost :/

can't i just take the fact that x = rcos$\theta$ and y= rsin$\theta$ and plug that into z = x +iy? because that'll give the desired results right?

 

[aaj92]

oh... then i still don't know how to get part b. k well I'll have to figure the whole Pythagorean theorem thing out then

 

[genericusrnme]

ok well I'm lost :/

can't i just take the fact that x = rcos$\theta$ and y= rsin$\theta$ and plug that into z = x +iy? because that'll give the desired results right?

You can but that isn't going to help you find r and $\theta$
I'll show you how to find r, then I'll let you try and find $\theta$

1. I'm going to square both of our equations to get

$x^2 = r^2 \ Cos(\theta )^2$
$y^2 = r^2 \ Sin(\theta )^2$

2. Next I'm going to add these equations together

$x^2 + y^2 = r^2 \ Cos(\theta )^2 + r^2 \ Sin(\theta )^2$

3. I'm going to pull out a common factor of $r^2$

$x^2 + y^2 = r^2 \ (Cos( \theta )^2 + Sin( \theta)^2 )$

4. I now use the fact that $Cos( \theta )^2 + Sin( \theta )^2 = 1$ to find

$x^2 + y^2 = r^2$

5. Taking the square root of both sides

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

Which as I said before gives us pythagoras' theorem


So in b) you have z = 3 + 4i, we can now find the corresponding r, $r = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

All that's left now is to find $\theta$

 

[aaj92]

oh my god! thank you! I didn't know you could just add them together sorry my brain is just refusing to work right now but yeah I see how you can get theta now. thank you so much :)

 

[genericusrnme]

No problem buddy!