Converting into rectangular form

Question

I have 2 related questions:

First:

Let $z_1 = 2+2i$ and $z_2 = 2-2i$. Find $z_1z_2 $ in rectangular form.

I have no idea...

I'm also clueless about this question:

Change the following to rectangular: $$r=\frac{9}{5-4\cos(\theta)}$$

The answer for the first one according to my answer key is $8$. And the answer for second one is

$$ 9x^2 - 72x + 25y^2 - 81 = 0 $$

Answer 1

Okay, we have two concepts going on that while related, are probably best learned independently, and then then the relationship can be explained.

$|z|$ is the distance $z$ is from the origin.

If $z = x+ iy, |z| = \sqrt {x^2 + y^2}$

If $z$ is in polar form $z = \rho (\cos \theta + i \sin \theta), |z| = \rho$

To find $|z_1z_2|$, you have choice, you can multiply them together, find the result, and then use the formula above to find the answer. However, it is nature of complex numbers that $|z_1z_2| = |z_1||z_2|$

Whoops, answering a different question... I am not going to delete this, because it is good to know. $z_1 = a + i\,b\\ z_2 = x + i\,y \\ z_1z_2 = (a + i\,b)(x + i\,y) = (ax + i\,bx + i\,ay + i^2\,by)$

$i^2 = -1$, and you can combine the $i$ terms.

$z_1z_2 = (ax - by) + i\,(ay + bx)$

If $z_1$ and $z_2$ are in polar...

$z_1 = \rho_1 (\cos \theta + i \sin\theta)\\ z_2 = \rho_2 (\cos \phi + i \sin\phi)\\ z_1z_2 =\rho_1\rho_2 (\cos\theta\cos\phi - \sin\theta \sin\phi + i(\sin\theta \cos \phi + \cos \theta \sin \phi)\\ z_1z_2 =\rho_1\rho_2 (\cos(\theta+\phi) + i\sin(\theta+\phi))$

Converting equations in polar form to rectangular, and vice versa... know this.

$x = r \cos \theta\\ y = r \sin \theta\\ r^2 = x^2 + y^2$

It is not an equation if you don't have an equals sign. What you have above is an expression with no context, and there is nothing to do with it...

$r = \frac{9}{5-4\cos\theta}$

Do some algebra to turn that $\cos \theta$ into an $r\cos\theta$ and then say $r\cos\theta = x.$

And, you will have a leftover $r,$ isolate it, square both sides, and $r^2 = x^2 + y^2$

~~~~~~

$r = \frac{9}{5-4\cos\theta}\\ r(5-4\cos\theta) = 9\\ 5r - 4r\cos\theta = 9\\ r\cos\theta = x\\ 5r - 4x = 9\\ 5r =9+4x\\ 25r^2 =(9+4x)^2\\ 25r^2 =81+72x + 16x^2\\ r^2 = x^2 + y^2\\ 25(x^2 + y^2) =81+72x + 16x^2\\ 9x^2 - 72 x + 25y^2 =81\\ $

Now, if you wanted to you could put it into standard form for an ellipse, but it doesn't sound that that is really necessary.

Answer 2

To multiply complex numbers, you just use the distributive property and the fact that $i^2=-1$. So $(2+2i)(2-2i)=2\cdot 2 + 2i \cdot 2 +2 \cdot (-2i)+2i \cdot (-2i)$ Can you keep going? Maybe you recognize the factorization of the difference of squares, which would make this example easier.

Answer 3

$$r=\frac9{5-4\cos(\theta)},\\ 5r-4r\cos(\theta)=5r-4x=9,\\ 25r^2=25(x^2+y^2)=(9+4x)^2=81+72x+16x^2.$$

Hence,

$$9x^2-72x+25y^2-81=0.$$

This is a cute ellipse.