Help with Complex Numbers

• Dec 5th 2012, 12:16 PM
INeedOfHelp
Help with Complex Numbers
1.Find the product of Attachment 26086 and Attachment 26087

Attachment 26089
Attachment 26090
Attachment 26091
Are those steps right so far? If so, how do I solve the rest?

2.If Attachment 26095 and Attachment 26093 find Attachment 26094

But the correct answer is 2+i sq.rt2
How do I get that answer?
• Dec 5th 2012, 01:10 PM
Plato
Re: Help with Complex Numbers
Quote:

Originally Posted by INeedOfHelp
1.Find the product of Attachment 26086 and Attachment 26087

Attachment 26089
Attachment 26090
Attachment 26091
Are those steps right so far? If so, how do I solve the rest?

2.If Attachment 26095 and Attachment 26093 find Attachment 26094

But the correct answer is 2+i sq.rt2
How do I get that answer?

I find your posting almost impossible to read. Why not learn LaTex?

Given any complex number, it is true $\displaystyle z\cdot\overline{z}=|z|^2$.
• Dec 5th 2012, 01:22 PM
HallsofIvy
Re: Help with Complex Numbers
Quote:

Originally Posted by INeedOfHelp
1.Find the product of Attachment 26086 and Attachment 26087

Attachment 26089
Attachment 26090
Attachment 26091
Are those steps right so far? If so, how do I solve the rest?

There are two ways of doing this. The first is essentially what you have done $\displaystyle sin(\pi/6)= \frac{1}{2}$ and $\displaystyle cos(\pi/6)= \frac{\sqrt{3}}{2}$ so that $\displaystyle z= \frac{3\sqrt{3}}{2}+ i\frac{3}{2}$. $\displaystyle \overline{z}= \frac{3\sqrt{3}}{2}- i\frac{3}{2}$. All that has to be multiplied. It is simpler if you realize that $\displaystyle (a+ bi)(a- bi)= a^2+ b^2$, a real number.

The other way is to use the exponential form. saying $\displaystyle z= 3(cos(\pi/6)+ i sin(\pi/6))$ is the same as saying $\displaystyle z= 3e^{i\pi/6}$. And then $\displaystyle \overline{z}= 3e^{-i\pi/6}$. And then the product is $\displaystyle (3)(3)(e^{i\pi/6- i\pi/6})= 9$

Quote:
$\displaystyle \frac{8e^{\frac{\pi i}{2}}}{4e^{\frac{\pi i}{4}}}= \frac{8}{4}e^{\frac{\pi i}{2}- \frac{\pi i}{4}}= 2e^{\frac{\pi i}{4}}$

Quote:

And, of course, $\displaystyle 2e^{\pi i/4}= 2(cos(\pi/4)+ i sin(\pi/4))$. What is that equal to? (It is NOT "2+ i sq.rt 2"! I suspect you have misread that.)