# Homework Help: Complex Number Question(s)

1. Jan 31, 2016

### King_Silver

1. The problem statement, all variables and given/known data
a) Solve equation z + 2i z(with a line above it i.e. complex conjugate) = -9 +2i
I want it in the form x + iy and I am solving for z.

b)
The equation |z-9+9i| = |z-6+3i| describes the straight line in the complex plane that is the perpendicular bisector of the line segment from 9-9i to 6-3i.
Find the value of m and c

2. Relevant equations
Complex conjugate: z = x + iy, z(line above it) = x -iy

3. The attempt at a solution
a) I'm not exactly sure how to approach this but I have a few ideas.
1. let z = -9 and 2i z(line above) = 2i and solve that way
2. Let the entire equation = z, however when I do this what do I do about the complex conjugate z?
3. Let entire equation = z complex conjugate and change the signs?
I really am unsure which will work.

b) If you write it in term of x +iy, you can write the equation like so: y = mx+c.
I now need to find the values of m and c, my textbook literally jumps to insanely easy complex numbers to questions like these, I presume they are just difficultly worded questions and that they aren't actually difficult. Any help on how to approach these questions?

2. Jan 31, 2016

### HallsofIvy

These are only "difficultly worded" if you don't understand what the words mean. What points in the plane represent the complex numbers 9- 9i and 6- 3i? What is the equation of the line through those two points? What is the equation of the perpendicular bisector of that line segment? $|x+ iy|= \sqrt{x^2+ y^2}$ so how can you write |z- 9- 9i|= |z-6+3i| in terms of x and y? What do you get when you square both sides of that?

Last edited by a moderator: Jan 31, 2016
3. Jan 31, 2016

### SteamKing

Staff Emeritus
Let's take a) which is relatively simple.

You need to solve $z + 2i\bar z = -9 +2i$

Use the definition of $z = x + iy$ and $\bar z = x - iy$ and substitute these back into the original equation. Do the multiplication of $\bar z$ and $2i$ as indicated.

You find z by equating the real part and the complex part on both sides of the equation.

4. Jan 31, 2016

### King_Silver

So rewrite it as... x+iy +2i x-iy = -9 +2i?

I understand what you mean by equating the real part and the complex part on both sides of the equation i.e. reals and reals together, imaginary and imaginary together right? :)
but what do you mean by "Do the multiplication of z(conjugate) +2i as indicated? :)

5. Jan 31, 2016

### SteamKing

Staff Emeritus
Exactly that. Calculate the product $2i ⋅ (x - iy)$. What do you get when you multiply everything out?

6. Jan 31, 2016

### HallsofIvy

No, you rewrite it as x+ iy- 2i(x- iy)= -9+ 2i. Do you see the difference?

He means "multiply the two complex numbers"!

7. Jan 31, 2016

### King_Silver

when you multiply both those together you get -2i2y but because i2 = -1 you get 2ix + 2? :/
Then you can rewrite it as follows right?
x+ iy- 2ix +2= -9+ 2i

8. Jan 31, 2016

### SteamKing

Staff Emeritus
What happened to the y when you multiplied 2i by -iy ?

9. Jan 31, 2016

### King_Silver

(2i)(x-iy)

(2i)(x) + (2i)(-iy)

2ix - 2i^2y (i^2 = -1)

2ix +2y whoops sorry. Is that better?

10. Jan 31, 2016

### SteamKing

Staff Emeritus
Yep.

11. Jan 31, 2016

### King_Silver

x + iy +2ix +2y = -9 + 2i

Now I want to put Re numbers with Re numbers and the Im with the Im numbers so...
x + 2y + 9 = -2ix -iy +2i

Am I making progress here? :)

12. Jan 31, 2016

### SteamKing

Staff Emeritus
A little, but it's all jumbled up.

Remember, you want to equate the real parts on the left side of the equation with the real parts on the other, namely the -9 + 2i quantity. Same for the complex parts.
You turn the original equation into two new equations in this process.

13. Jan 31, 2016

### King_Silver

So in other words, keep it like in this form: x + iy +2ix +2y = -9 + 2i
Then equate Re with Re and Im with Im so for example, the Real parts in this case...

x + 2y = -9 ?

14. Jan 31, 2016

### SteamKing

Staff Emeritus
Yep. And do the same for the imaginary parts. You should wind up with two equations in two unknowns. Solve for x and y.

15. Jan 31, 2016

### King_Silver

x+2y = -9
2x + y = 2

Solve simultaneously and I should get the value of x and y to be fractions, then it will look something like x = a/b, y = c/di
z = a/b + c/di where a,b,c,d represent numbers? :)

16. Jan 31, 2016

### SteamKing

Staff Emeritus
Yes. And you can check your calculations by substituting the values of x and y back into the original equation.

17. Jan 31, 2016

### King_Silver

Woo :D I understand now! thanks!! huge help! now for Question b.
As HallsofIvy said above, |z-9+9i| can be written in the form √x^2 + y^2. However in |z-9+9i| I have 3 variables, a z, a real number and an imaginary number. do I just replace the z with x+iy again in this case?

18. Jan 31, 2016

### SteamKing

Staff Emeritus
Yes. The definition z = x + iy can always be used.

19. Feb 1, 2016

### King_Silver

Sorry for the late response I was studying.
|z-9+9i| = |x+iy-9+9i|
|z-6+3i| = |x+iy-6+3i|
|x+iy-9+9i| = |x+iy-6+3i| or √(-9+x)^2 + (9i +iy)^2 = √(-6+x)^2 + (3i+iy)^2

Is this correct?

20. Feb 1, 2016

### King_Silver

21. Feb 1, 2016

### SteamKing

Staff Emeritus
Not quite.

When calculating the absolute value |z|, that is defined as |z| = |x + iy| = $\sqrt{x^2 + y^2}$

Note that only the y value is used in the calculation, not iy.

22. Feb 1, 2016

### King_Silver

so |z-9+9i| would be rewritten as √9^2+9^2 ? or as √x^2 + y^2 , +9+9i

23. Feb 1, 2016

### SteamKing

Staff Emeritus
No. If the complex number w = z - 9 + 9i, then in order to write |w|, we collect real and imaginary terms separately first.

Using the definition z = x + iy, then w = (x - 9) + i * (y + 9), so that $|w| = \sqrt{(x-9)^2+ (y+9)^2}$, according to the definition of absolute value for complex numbers.

24. Feb 1, 2016

### King_Silver

Ok that makes more sense now.
So for
w = z -9+9i, |w| = |x+iy -9 +9i| or |w| = √(x-9)^2 + (y+9)^2
w = z -6+3i, |w| = |x+iy -6 +3i| or |w| = √(x-6)^2 + (y+3)^2

Both of these are equal each other. so...
√(x-9)^2 + (y+9)^2 = √(x-6)^2 + (y+3)^2
√x^2 -18x+81 +y^2 +18y + 81 = √x^2-12x+36 + y^2 +6y+9

And to get rid of the square roots you square both sides right?
then gather ys together and xs together, solve for x = 0 to find y's value and y = 0 to solve for x's value
sub them in then?

25. Feb 1, 2016

### SteamKing

Staff Emeritus
Yes.
You'll wind up with a quadratic in x = a quadratic in y, after collecting terms.

I think for part b) of the original question, you were supposed to be calculating m and c for a straight line. Don't lose sight of that.