# Hi, having problems with a complex number multiplication

• Mar 22nd 2013, 10:13 AM
Jojo55
Hi, having problems with a complex number multiplication
Hi everyone, I'm Joanne and currently studying to become a maths teacher (1 year in to studying).

I'm having problems trying to work out the below multiplication.

Attachment 27644

I've worked out that the polar form of (2-2i)^12 is Attachment 27645 ( sorry about the ?'s)

But I am unsure of what to do next.

Do I work out the polar form of Attachment 27646, then multiply them together, or am I doing this wrong to begin with?

Thanks
• Mar 22nd 2013, 10:26 AM
Plato
Re: Hi, having problems with a complex number multiplication
Quote:

Originally Posted by Jojo55
Hi everyone, I'm Joanne and currently studying to become a maths teacher (1 year in to studying).
I'm having problems trying to work out the below multiplication.
Attachment 27644
I've worked out that the polar form of (2-2i)^12 is Attachment 27645 ( sorry about the ?'s)

$(2-2i)=2\sqrt 2 \exp \left( {\frac{{ -i \pi }}{4}} \right)$.

thus $(2-2i)^{12}=\left(2\sqrt 2\right)^{12} \exp \left( -3i\pi \right)$.

But $\left(2\sqrt 2\right)^{12}=2^{18}$.

Can you finish?
• Mar 22nd 2013, 11:31 AM
jpritch422
Re: Hi, having problems with a complex number multiplication
Are you familiar with the Binomial Theorem?

• Mar 22nd 2013, 12:03 PM
Plato
Re: Hi, having problems with a complex number multiplication
Quote:

Originally Posted by jpritch422
Are you familiar with the Binomial Theorem?

What does that have to do with the OP?
• Mar 22nd 2013, 01:15 PM
Henderson
Re: Hi, having problems with a complex number multiplication
Quote:

Originally Posted by Plato
What does that have to do with the OP?

I think jpritch is suggesting that the problem be done without DeMoivre/Polar form.
I disagree with that suggestion, but I'm fairly certain that's where he's going with that.
• Mar 22nd 2013, 02:08 PM
Plato
Re: Hi, having problems with a complex number multiplication
Quote:

Originally Posted by Henderson
I think jpritch is suggesting that the problem be done without DeMoivre/Polar form.
I disagree with that suggestion, but I'm fairly certain that's where he's going with that.

Actually, I understood what the post meant. I want to see why jpritch422 wanted to use that.

The irony is that DeMoivre's Rule is used for many proofs about binomial coefficients.
• Mar 25th 2013, 05:13 AM
Jojo55
Re: Hi, having problems with a complex number multiplication
Quote:

Originally Posted by Plato
$(2-2i)=2\sqrt 2 \exp \left( {\frac{{ -i \pi }}{4}} \right)$.

thus $(2-2i)^{12}=\left(2\sqrt 2\right)^{12} \exp \left( -3i\pi \right)$.

But $\left(2\sqrt 2\right)^{12}=2^{18}$.

Can you finish?

Sorry for the late reply. I managed to solve to get the answer of -1

Thanks for the help on this