# Homework Help: Complex Numbers

1. Sep 1, 2012

### GreenPrint

1. The problem statement, all variables and given/known data

Evaluate (find the real and complex components) of the following complex numbers, in either rectangular or polar form:

$z_{1}$ = $\frac{j(3-j4)^{*}}{(-1+6j)(2+j)^{2}}$

2. Relevant equations

$e^{jθ}$ = cosθ + j sinθ

3. The attempt at a solution

I sadly don't even know where to begin here. I understand that my textbook uses j instead of i for imaginary number. I understand that a star superscript means the complex conjurgate so...

$z_{1}$ = $\frac{j(3-j4)^{*}}{(-1+6j)(2+j)^{2}}$ = $\frac{j(j4-3)}{(-1+6j)(2+j)^{2}}$

Is this true? Where do I go from here thanks.

Just a note.

The prerequisite for this course was an introductory differential equations course, multivariable calculus and a second semester of physics, all of which I have taken. I still have no idea what half of this stuff is. I've never studied complex numbers before but dove into euler's formula and stuff of the like in high school out of curiosity. I however am not sure what a phasor is exactly but have some understand of the concept. Thanks for any help or suggestions on how to proceed to solve this problem thanks.

Last edited: Sep 1, 2012
2. Sep 1, 2012

### micromass

The first thing you can do is work out all the brackets. What do you get once you do that?

3. Sep 1, 2012

### Dick

Your first step is a little off. The complex conjugate of 3-j4 is 3+j4. Then you can just start multiplying the numerator and denominator out using j*j=(-1). Once you've got something in the form (a+bj)/(c+dj) you multiply numerator and denominator by the complex conjugate of (c+dj) to make the denominator real. There's nothing really subtle involved. It's just a lot of arithmetic.

4. Sep 1, 2012

### GreenPrint

$\frac{4-3j}{106j-41}$

oh ok let me see here

5. Sep 1, 2012

### Ray Vickson

NO: $(3 - 4\,j)^{*} \neq -3 + 4\, j;$ taking the complex conjugate changes the sign of the imaginary part only, and does not affect the real part.

Anyway: express the whole numerator in the form $a + j \, b$ for real 'a' and 'b', and express the whole denominator in the form $c + j \, d$ for real 'c' and 'd'. Then use the standard quotient rule to evaluate
$$\frac{a + j\,b}{c + j\, d}$$
in whatever final form you choose (either as $A + j\, B$ or as $r\, e^{j \theta}$). Basically, that is how such questions are always done.

RGV

6. Sep 1, 2012

### GreenPrint

$\frac{266+371j}{9555}$

where do i go from here?

7. Sep 1, 2012

### Dick

I have no idea what you are doing, but those numbers don't look anything like what I'm getting. Can you spell out your calculation in detail?

8. Sep 1, 2012

### GreenPrint

Ya I think I screwed it up I get

2/315 + 47/315 j

I hope that is correct. There's nothing left to do at this point?

9. Sep 1, 2012

### Dick

Nothing left to do except try and get the numbers right. I still don't agree with you. What did you get for the numerator and the denominator?

10. Sep 1, 2012

### GreenPrint

What

$\frac{{(3-4j)}^{*}j}{(-1+6j)(2+j)^{2}}$

${(3-4j)}^{*}=(3+4j)$
${(2+j)}^{2}=2^{2}+j^{2}+2(2)j=4+1+4j=5+4j$

$\frac{(3+4j)j}{(-1+6j)(5+4j)}$

$(3+4j)j=3j+4j^{2}=3j+4$
$(-1+6j)(5+4j)=-5-4j+30j+24j^{2}=-5+26j+24=19+26j$

$\frac{3j+4}{19+26j}$

$\frac{3j+4}{19+26j}*\frac{19-26j}{19-26j}$

(19+26j)(19-26j)=$19^{2}-26(19)j+26(19)j-26^{2}j^{2} = 19^{2} - 26^{2} = 361 - 676 = -315$

$(3j+4)(19-26j)=3(19)j-3(26)j^{2}+4(19)-26(4)j=57j-78+76-104j = -2-47j$

$\frac{-2-47j}{-351} = \frac{2}{315}+\frac{47j}{315}$

don't see what i did wrong

11. Sep 1, 2012

### micromass

You did $j^2=1$. It should be $j^2=-1$.

12. Sep 1, 2012

### Dick

Um, j^2=(-1). Not +1. Check (2+j)^2 again. It's not 5+4j, is it? Etc.

13. Sep 1, 2012

### GreenPrint

6/37 - 1/37 j ?

14. Sep 1, 2012

### Dick

That's what I get.

15. Sep 2, 2012

### GreenPrint

Thank you much

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