Simple complex number problem: z=x+iy, z1=2-3i, z konjugated = x-iy

[Marko1337]

Hey guys i need help solving this.

z=x+iy
z1=2-3i
z konjugated = x-iy


Real part of (z*z1) = 18
Imaginary part of ((z konjugated)/z1)=1

I need to calculate the values of x and y

Please help, i will appreciate it a lot

 

[Dr.Peterson]

You have not shown any work, or any information from which we can determine what kind of help you need. We are not here to do your work for you, but to help you learn how to solve it.

Please be specific about what you don't understand or where you got stuck in solving the problem. The more work you show, the more effectively we can help.

Each of the two conditions will yield a linear equation in x and y, which you can then solve.

 

[Marko1337]

You have not shown any work, or any information from which we can determine what kind of help you need. We are not here to do your work for you, but to help you learn how to solve it.

Please be specific about what you don't understand or where you got stuck in solving the problem. The more work you show, the more effectively we can help.

Each of the two conditions will yield a linear equation in x and y, which you can then solve.

My bad for forgeting to say what ive manage to calculate, sorry about that, i did work on it for a couple hours. And all ive been able to do is to calculate this :

Real part of (z*z1) = 2x + 3y = 18
Imaginary part of ((z konjugated)/z1)= 3x*i - 2y*i

This is where im stuck, i do not know how to calculate x and y from here

 

[Dr.Peterson]

My bad for forgeting to say what ive manage to calculate, sorry about that, i did work on it for a couple hours. And all ive been able to do is to calculate this :

Real part of (z*z1) = 2x + 3y = 18
Imaginary part of ((z konjugated)/z1)= 3x*i - 2y*i

This is where im stuck, i do not know how to calculate x and y from here

The problem was:

z=x+iy
z1=2-3i
z conjugated = x-iy
Real part of (z*z1) = 18
Imaginary part of ((z conjugated)/z1)=1

You have found that \(\displaystyle Re(zz_1) = Re((x+iy)(2-3i)) = 2x+3y\), so the first equation is 2x+3y = 18. That's good.

But it appears that you have misinterpreted "imaginary part". The imaginary part of a+ib is just b, not ib; if you think of a+ib as the ordered pair (a,b), the imaginary part is the coordinate b, not the imaginary number ib.

So \(\displaystyle Im(\dfrac{\overline{z}}{z_1})\) is not 3x*i - 2y*i, but just 3x-2y. So the second equation is 3x-2y = 1.

Now just solve the system of equations.

But that's not quite right; \(\displaystyle Im(\dfrac{\overline{z}}{z_1})\) is not really 3x-2y. You appear to have omitted the denominator. That will change the second equation.

 

[Marko1337]

The problem was:


You have found that \(\displaystyle Re(zz_1) = Re((x+iy)(2-3i)) = 2x+3y\), so the first equation is 2x+3y = 18. That's good.

But it appears that you have misinterpreted "imaginary part". The imaginary part of a+ib is just b, not ib; if you think of a+ib as the ordered pair (a,b), the imaginary part is the coordinate b, not the imaginary number ib.

So \(\displaystyle Im(\dfrac{\overline{z}}{z_1})\) is not 3x*i - 2y*i, but just 3x-2y. So the second equation is 3x-2y = 1.

Now just solve the system of equations.

But that's not quite right; \(\displaystyle Im(\dfrac{\overline{z}}{z_1})\) is not really 3x-2y. You appear to have omitted the denominator. That will change the second equation.

Okay thank you very much. Going right back to solving it