Help with Complex Numbers

**INeedOfHelp** · December 5th 2012, 12:16 PM

1.Find the product of

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and

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Help with Complex Numbers-codecogseqn-6-.gif

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Help with Complex Numbers-codecogseqn-8-.gif

Are those steps right so far? If so, how do I solve the rest?

2.If

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and

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find

Help with Complex Numbers-codecogseqn-11-.gif

I got the answer 2e^(2pi/8)i

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

**Plato** · December 5th 2012, 01:10 PM

Originally Posted by INeedOfHelp

1.Find the product of

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and

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Are those steps right so far? If so, how do I solve the rest?

2.If

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and

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find

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I got the answer 2e^(2pi/8)i

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 $z\cdot\overline{z}=|z|^2$ .

**HallsofIvy** · December 5th 2012, 01:22 PM

Originally Posted by INeedOfHelp

1.Find the product of

and

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 $sin(\pi/6)= \frac{1}{2}$ and $cos(\pi/6)= \frac{\sqrt{3}}{2}$ so that $z= \frac{3\sqrt{3}}{2}+ i\frac{3}{2}$ . $\overline{z}= \frac{3\sqrt{3}}{2}- i\frac{3}{2}$ . All that has to be multiplied. It is simpler if you realize that $(a+ bi)(a- bi)= a^2+ b^2$ , a real number.

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

2.If

and

find

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

I got the answer 2e^(2pi/8)i

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

First if you do not know that 2/8= 1/4 you should give up all math!

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

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