Complex Number Equations: Solving for z and Finding the Perpendicular Bisector

[King_Silver]

Homework Statement

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

Homework Equations

Complex conjugate: z = x + iy, z(line above it) = x -iy

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?

 

[HallsofIvy]

Homework Statement

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

Homework Equations

Complex conjugate: z = x + iy, z(line above it) = x -iy

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

You are told that the product of z+ 2i and its conjugate is -9+ 2i. Why would you assume z itself is equal to that?
Let z= x+ iy. Then z+ 2i= x+ (y+ 2)i and its conjugate is x- yi. What is the product of those? Set the real part of the product equal to -9, the imaginary part equal to 2 and solve the two equations for x and y.
(I see that Steam King is interpreting this as $z+ 2i(\overline{z})$ rather than $(z+ 2i)\overline{z}$ as I did.)

Let the entire equation = z, however when I do this what do I do about the complex conjugate z?

1. 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?

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?

 

[SteamKing]

Homework Statement

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

Homework Equations

Complex conjugate: z = x + iy, z(line above it) = x -iy

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?

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.

 

[King_Silver]

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.

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? :)

 

[SteamKing]

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? :)

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

 

[HallsofIvy]

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

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

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? :)

He means "multiply the two complex numbers"!

 

[King_Silver]

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

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

 

[SteamKing]

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

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

 

[King_Silver]

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

(2i)(x-iy)

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

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

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

 

[SteamKing]

(2i)(x-iy)

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

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

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

Yep.

 

[King_Silver]

Yep.

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

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

Am I making progress here? :)

 

[SteamKing]

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

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

Am I making progress here? :)

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.

 

[King_Silver]

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.

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

x + 2y = -9 ?

 

[SteamKing]

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

x + 2y = -9 ?

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

 

[King_Silver]

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

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? :)

 

[SteamKing]

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? :)

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

 

[King_Silver]

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

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?

 

[SteamKing]

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?

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

 

[King_Silver]

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

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?

 

[King_Silver]

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

Why did the +2i become a -2i actually? what is the reasoning behind this?

 

[SteamKing]

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?

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.

 

[King_Silver]

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.

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

 

[SteamKing]

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

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.

 

[King_Silver]

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.

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?

 

[SteamKing]

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?

Yes.

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?

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.

 

[King_Silver]

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.

Yea I have to find y =mx+c but I need to find out how to obtain the values of m and the values of c.

Would letting x = 0 be of any help for that?
like y = m(0) + c. That way m would cancel and c = y.

 

[Ray Vickson]

Yea I have to find y =mx+c but I need to find out how to obtain the values of m and the values of c.

Would letting x = 0 be of any help for that?
like y = m(0) + c. That way m would cancel and c = y.

Why not try it out for yourself, to see if it works?

 

[King_Silver]

Why not try it out for yourself, to see if it works?

I would, only problem is the workspace I am working with is horrible, its an online question generator my lecturer uses.
Pretty much he puts up the questions for practice, you enter your answer only and then submit but only after submitting can you know if it is right or wrong.
If it happens to be wrong, I need to restart EVERY question from scratch over and over due to the random number generator. It is a pain in the ass so I want to know are my methods correct before I keep submitting the same nonsense over and over again without actually learning anything.

 

[Ray Vickson]

I would, only problem is the workspace I am working with is horrible, its an online question generator my lecturer uses.
Pretty much he puts up the questions for practice, you enter your answer only and then submit but only after submitting can you know if it is right or wrong.
If it happens to be wrong, I need to restart EVERY question from scratch over and over due to the random number generator. It is a pain in the ass so I want to know are my methods correct before I keep submitting the same nonsense over and over again without actually learning anything.

Can't you use scrap paper first, then copy out the solution?

 

[King_Silver]

For Question A of the original post I got: -13/3+20/3*i
For Question

Can't you use scrap paper first, then copy out the solution?

Will I write out my questions on paper, my attempt at the solutions alongside them. I think that would be more beneficial for myself and the people trying to help me, I don't think I am doing myself any favours at the moment. I'll write out the questions there and my attempts at them :) be back in a bit

 

[King_Silver]

Here are my 3 Questions so far:
Q1 and Q3 on the images represent the 2 I asked in the OP. The Q2 is one I have done myself and I believe to be right. Is this any better?

 

[King_Silver]

Here are my 3 Questions so far:
Q1 and Q3 on the images represent the 2 I asked in the OP. The Q2 is one I have done myself and I believe to be right. Is this any better?

any help? :)