Help Solving Equation 2iz-3conjugate[z]=-12+13i

[luff]

I've tried for a while to solve the this equation but I can't seem to get it right.

2iz-3conjugate[z]=-12+13i

Anyone who could help me tackle this one?

cheers

 

[stapel]

I've tried for a while to solve the this equation but I can't seem to get it right.

2iz-3conjugate[z]=-12+13i

We'll be glad to help you find where things are going wrong, but we'll need to see what you've tried, first. So please provide a clear listing of your steps so far, including any information you may have been given about "z". Thank you!

 

[pka]

I've tried for a while to solve the this equation but I can't seem to get it right.
2iz-3conjugate[z]=-12+13i

You need to try to solve this and post the effort. Then we can see where you need help.

Can you solve \(\displaystyle \left\{ \begin{array}{l} 3x + 2y = 12\\ 2x + 3y = 13 \end{array} \right.\;?\) What do those have to do with this question?

 

[luff]

So far I've tried to solve out conjugate[z] by dividing -12 + 13i with 2iz.
So (-12+13i)/2iz and then multiplying with the conjugate on both sides:
((-12+13i)*-2iz)/(2iz*-2iz) = (24iz + 26z)/ 4z^2 = -3conjugate[z]

which should lead to (24i + 26)z / (-12z)z = Conjugate[z] = (-2i - 3) / z

 

[stapel]

So far I've tried to solve out conjugate[z] by dividing -12 + 13i with 2iz.

This would lead to:

. . . . .\(\displaystyle \dfrac{2i\,z\, -\, 3\,\overline{z}}{2i\,z}\, =\, \dfrac{-12\, +\, 13i}{2i\,z}\)

. . . . .\(\displaystyle 1\, -\, \dfrac{3\,\overline{z}}{2i\,z}\, =\, -\dfrac{6}{i\,z}\, +\, \dfrac{13}{2\,z}\)

So (-12+13i)/2iz and then multiplying with the conjugate on both sides:

This would lead to:

. . . . .\(\displaystyle -2i\,z\, +\, 3\, \overline{z}\, =\, 12\, -\, 13i\)

((-12+13i)*-2iz)/(2iz*-2iz) = (24iz + 26z)/ 4z^2 = -3conjugate[z]

which should lead to (24i + 26)z / (-12z)z = Conjugate[z] = (-2i - 3) / z

I'm sorry, but I don't follow. Please reply showing your steps.

Note: The hints and helps you've been provided before might be more useful. For instance, letting z = a + bi, then the conjugate of z is a - bi. Then we start with:

. . . . .\(\displaystyle 2i\, (a\, +\, bi)\, -\, 3\, (a\, -\, bi)\, =\, -12\, +\, 13i\)

. . . . .\(\displaystyle 2ai\, -\, 2b\, -\, 3a\, +\, 3bi\, =\, -12\, +\, 13i\)

. . . . .\(\displaystyle (-2b - 3a) + (2a + 3b)\, i\, =\, -12\, +\, 13i\)

You have chosen not to use this method. Did the instructions for this exercise specific that you had to use some other method? Thank you.

 

[pka]

which should lead to (24i + 26)z / (-12z)z = Conjugate[z] = (-2i - 3) / z

That can never be the case. i.e. \(\displaystyle \overline{~z~}\not=\dfrac{-2i-3}{z}\).
The reason being that \(\displaystyle z\cdot\overline{~z~}=|z|^2\), which is a real number.

Frankly, I do not think that this can be solved with that approach.

As I did in my first reply, let \(\displaystyle z=x+yi\).
Then there will be two equations in two variables to solve.

 

[luff]

Saw my mistakes in my earlier attempts and will look into the way you suggested. Thanks