Complex numbers problem |z| - iz = 1-2i

[Callmelucky]

Homework Statement

Photo of procedure below.

Relevant Equations

z= x+yi, |z|=sqrt(x^2 + y^2), i^2=-1

Here is my attempt(photo below), but somehow the solution in the textbook is z= 2 - (3/2)i, and mine is z=(-3/2) +2i.

Can someone please tell me where I am making a mistake? I suppose it's something with x being part of the real part of the 1st complex number and x being part of an imaginary number of the 1st complex number. But I don't know how to make it right.

Thank you.

 

[SammyS]

Homework Statement:: Photo of procedure below.
Relevant Equations:: z= x+yi, |z|=sqrt(x^2 + y^2), i^2=-1

Here is my attempt(photo below), but somehow the solution in the textbook is z= 2 - (3/2)i, and mine is z=(-2/3) +2i.

Can someone please tell me where I am making a mistake? I suppose it's something with x being part of the real part of the 1st complex number and x being part of an imaginary number of the 1st complex number. But I don't know how to make it right.

Thank you.

Please include the statement of the problem in the text of your post, not merely in the Title.

Also, the image you posted is unreadable or nearly so.

Please learn to use the typing helps at the top of the editing screen, at the very least. Better yet, learn to use LaTeX.

 

[Callmelucky]

Please include the statement of the problem in the text of your post, not merely in the Title.

Also, the image you posted is unreadable or nearly so.

Please learn to use the typing helps at the top of the editing screen, at the very least. Better yet, learn to use LaTeX.

I want to learn latex but I don't have time right now. Here is another pic
Here is the new pic(below). I want to learn Latex, but I don't have time rn.
Problem: Solve in the set of complex numbers. |z| -iz = 1-2i

 

[Mark44]

Please include the statement of the problem in the text of your post, not merely in the Title.

Also, the image you posted is unreadable or nearly so.

Please learn to use the typing helps at the top of the editing screen, at the very least. Better yet, learn to use LaTeX.

Amen to all of these. As @SammyS noted, the image is pretty much unreadable. However, I was able to work the problem and get the same solution as shown in the book.

I suppose it's something with x being part of the real part of the 1st complex number and x being part of an imaginary number of the 1st complex number.

This is very unclear.
The equation you're working with is $|z| - iz = 1 - 2i$
Replace z with x + iy and substitute into the given equation. Note that in the complex number x + iy, y is the coefficient of the imaginary part, but both x and y are real numbers.

 

[Mark44]

The new image is slightly more readable, and it looks like you were able to correct your mistake.
Edit: I didn't notice that the real and imaginary parts were reversed.

 

[Callmelucky]

Yeah but I don't understand how is y imaginary part if i is next to x(xi). Shouldn't x be imaginary part?

 

[Callmelucky]

I know that z= x+yi, but I have xi after multiplying i(x+yi). That is what bothers me.

 

[PeroK]

I know that z= x+yi, but I have xi after multiplying i(x+yi). That is what bothers me.

If $z = x +yi$, then $iz = -y +ix$ and we see that $Im(iz) = x$.

 

[DaveE]

OK, so; you said:
1) $z=x + iy$
2) $x=2$
3) $y=-\frac{3}{2}$
and (drum roll, please)...
4) $z=-\frac{3}{2} + 2i$

Your only problem is your last line.

 

[Callmelucky]

OK, so; you said:
1) $z=x + iy$
2) $x=2$
3) $y=-\frac{3}{2}$
and (drum roll, please)...
4) $z=-\frac{3}{2} + 2i$

Your only problem is the last line.

But that is not the solution. Solution is z= 2 - (3/2)i

 

[DaveE]

But that is not the solution. Solution is z= 2 - (3/2)i

Reread my post slowly. Do the substitutions in step 4 carefully.

 

[Callmelucky]

Reread my post slowly. Do the substitutions in step 4 carefully.

I don't get how did x became y and vice versa

 

[PeroK]

But that is not the solution. Solution is z= 2 - (3/2)i

Let's assume your solution is correct:
$$z = -\frac 3 2 + 2i$$$$\Rightarrow \ |z| = \frac 5 2, iz = -2 - \frac 3 2i$$$$\Rightarrow \ |z| - iz = \frac 9 2 + \frac 3 2 i$$

 

[DaveE]

I don't get how did x became y and vice versa

Me either.

 

[PeroK]

I don't get how did x became y and vice versa

Do you recognise that $z$ and $iz$ are different complex numbers? In the sense that $z \ne iz$?

 

[Callmelucky]

Do you recognise that $z$ and $iz$ are different complex numbers? In the sense that $z \ne iz$?

I didn't recognize that. What does that mean, that I have 3 complex numbers in that equation?

 

[PeroK]

I didn't recognize that. What does that mean, that I have 3 complex numbers in that equation?

It means that if $z = x + iy$ then $iz \ne x + iy$.

 

[Callmelucky]

It means that if $z = x + iy$ then $iz \ne x + iy$.

I got the right solution. I think I get it now. For some reason I thought that I am not allowed to move stuff from right side to the left. Thank you.

 

[Callmelucky]

Thanks everyone.

 

[FactChecker]

I don't get how did x became y and vice versa

The $x$ and $y$ came from $z=x+iy$ (which you never specifically wrote down). So $x=2$ and $y = -3/2$ means that $z = 2 - (3/2) i$

 

[Mark44]

Yeah but I don't understand how is y imaginary part if i is next to x(xi). Shouldn't x be imaginary part?

If z = x + iy, then the Re(z) = x and Im(z) = y.

I know that z= x+yi, but I have xi after multiplying i(x+yi).

$iz = i(x + iy) = i^2y + ix = -y + ix$
Re(iz) = -y and Im(iz) = x
The real part is the part that isn't multiplied by i. The imaginary part is the coefficient of i.

I don't get how did x became y and vice versa

It didn't. What you missed is that the Re(al) and Im(aginary) parts got changed after multiplication by i.