How to use DeMoivre's theorem to find the product of complex numbers?

[karush]

Find product
(1+3i)(2-2i)
then change each to complex form
and find product
with DeMoivres theorem

clueless!

 

[Greg]

$$q=\arctan(3)$$

$$w=\arctan(-1)$$

$$\sqrt{10}(\cos(q)+i\sin(q))\cdot\sqrt{8}(\cos(w)+i\sin(w))=8+4i$$

 

[MarkFL]

I'm wondering if what we're supposed to do here is:

$$1+3i=(1+i)(2+i)$$

$$2-2i=-(1+i)^3$$

And so the product is:

$$(1+3i)(2-2i)=-(1+i)^4(2+i)=-(1+i)^4(1+(1+i))=-\left((1+i)^4+(1+i)^5\right)$$

Now, we find:

$$1+i=\sqrt{2}\cis\left(\frac{\pi}{4}\right)$$

And so, applying de Moivre's theorem, we have:

$$(1+3i)(2-2i)=-4\left(\cis\left(\frac{4\pi}{4}\right)+\sqrt{2}\cis\left(\frac{5\pi}{4}\right)\right)=-4\left((-1+0i)-\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)\right)=4(2+i)=8+4i$$

 

[karush]

im pretty sure that's what we are supposed to do

im just reviewing ahead for a class but this is all that was given.

MHB is my first choice for help...

 

[MarkFL]

im pretty sure that's what we are supposed to do

im just reviewing ahead for a class but this is all that was given.

MHB is my first choice for help...

What definition for "complex form" were you given? That was the confusing issue for me. I am aware of rectangular, polar and exponential forms for complex numbers, but I don't know what is meant by complex form. :)

 

[karush]

https://www.physicsforums.com/attachments/7187

this was it!

 

[MarkFL]

In case you're wondering how I obtained the following factorizations:

$$1+3i=(1+i)(2+i)$$

$$2-2i=-(1+i)^3$$

What I did for the first was to write:

$$(1+3i)(1-3i)=1-9i^2=10=5\cdot2=\left(2^2+1^2\right)\left(1^2+1^2\right)=(2+i)(2-i)(1+i)(1-i)$$

From there, it was easy to see that:

$$1+3i=(1+i)(2+i)$$

And for the second I wrote:

$$(2-2i)(2+2i)=8=2^3=(1^2+1^2)^3=(1+i)^3(1-i)^3$$

And then I computed:

$$(1+i)^3=1^3+3\cdot1^2i+3\cdot1i^2+i^3=1+3i-3-i=-2+2i=-(2-2i)\implies 2-2i=-(1+i)^3$$

 

[karush]

I'm wondering if what we're supposed to do here is:

$$1+3i=(1+i)(2+i)$$

$$2-2i=-(1+i)^3$$

And so the product is:

$$(1+3i)(2-2i)=-(1+i)^4(2+i)=-(1+i)^4(1+(1+i))=-\left((1+i)^4+(1+i)^5\right)$$

Now, we find:

$$1+i=\sqrt{2}\cis\left(\frac{\pi}{4}\right)$$

And so, applying de Moivre's theorem, we have:

$$(1+3i)(2-2i)=-4\left(\cis\left(\frac{4\pi}{4}\right)+\sqrt{2}\cis\left(\frac{5\pi}{4}\right)\right)=-4\left((-1+0i)-\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)\right)=4(2+i)=8+4i$$

by cis do you mean $\cos$

 

[I like Serena]

by cis do you mean $\cos$

$\cis$ means:
$$\cis\phi=e^{i\phi}=\cos\phi + i\sin\phi$$
It's a shorthand.

 

[karush]

$\cis$ means:
$$\cis\phi=e^{i\phi}=\cos\phi + i\sin\phi$$
It's a shorthand.

ok its not in the textbook!

$e^{i\phi}$ wasn't either!

 

[I like Serena]

ok its not in the textbook!

$e^{i\phi}$ wasn't either!

Sorry for running ahead on the material you're learning. (Blush)

 

[karush]

its fine
very useful point!

don't really think DeMoivre's is used very much
maybe down the road

 

[karush]

OK this is for my collection of solutions (as gratefully plagiarized};);)
I'm confident that there may be some oops in it..:p

$\textit{Obtain the following factorizations:}$
\begin{align*}\displaystyle
1+3i&=(1+i)(2+i)\\
2-2i&=-(1+i)^3
\end{align*}
$\textit{Then}$
\begin{align*}\displaystyle
(1+3i)(1-3i)&=1-9i^2=10=5\cdot2\\
&=\left(2^2+1^2\right)\left(1^2+1^2\right)\\
&=(2+i)(2-i)(1+i)(1-i)
\end{align*}
\begin{align*}\displaystyle
(1+3i)(2-2i)&=-(1+i)^4(2+i)\\
&=-(1+i)^4(1+(1+i))\\
&=-\left((1+i)^4+(1+i)^5\right)
\end{align*}
$\textsf{b. change each to complex form and find product. (with DeMoine's Theorem)}$
\begin{align*}\displaystyle
(1+3i)(1-3i)&=-4
\left[\left[
\cos\left(\frac{\pi}{4}\right)
+\sin\left(\frac{\pi}{4}\right)i
\right]
+\sqrt{2}
\left[\cos\left(\frac{5\pi}{4}\right)
+\sin\left(\frac{5\pi}{4}\right)i
\right]\right]\\
&=-4\left[(-1+0i)
-\sqrt{2}
\left(
\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}i
\right) \right]\\
&=4(2+i)\\
&=8+4i
\end{align*}

 

[Greg]

Brilliant solution, Mark. I like to avoid such a level of involvement but yours is very instructive. (Yes)

Nice thread.