Solve Complex Numbers: $z$ | Arg & Modulus Equations

[DrunkenOldFool]

Solve for complex number, $z$:\[\text{arg}\left( \frac{3z-6-3i}{2z-8-6i}\right)=\frac{\pi}{4}\]and \[|z-3+i|=3\]The problem I am facing is that when I substitute $z=x+iy$, the equations become extremely complicated. There has to be another tricky method which I am not able to figure out.

 

[CaptainBlack]

Solve for complex number, $z$:\[\text{arg}\left( \frac{3z-6-3i}{2z-8-6i}\right)=\frac{\pi}{4}\]and \[|z-3+i|=3\]The problem I am facing is that when I substitute $z=x+iy$, the equations become extremely complicated. There has to be another tricky method which I am not able to figure out.

The following may help:

The second condition means that \(z\) is a point on a circle of radius 3 centred at \(3-i\)

Also \(\arg(u/v)=(\arg(u)-\arg(v)) \mod 2\pi\).

CB

 

[Opalg]

Notice first that $\arg\left( \dfrac{3z-6-3i}{2z-8-6i}\right) = \arg\left( \dfrac32\,\dfrac{z-2-i}{z-4-3i}\right) = \arg\left( \dfrac{z-2-i}{z-4-3i}\right)$ (because multiplying a complex number by a positive real number does not alter its arg).

Next, $\arg\left( \dfrac{z-2-i}{z-4-3i}\right) = \arg\bigl(z-(2+i)\bigr) - \arg\bigl(z-(4+3i)\bigr)$. Now $z-(2+i)$ is the vector from $z$ to $2+i$, and $z-(4+3i)$ is the vector from $z$ to $4+3i$ (see the dashed lines in the picture below). We want the angle between those two vectors to be $\pi/4$.

If you recall your euclidean geometry (theorems about angles in the same segment, and the angle at the centre being twice the angle at the circumference), you will see that this requires $z$ to lie on a circle through the two black dots in the picture (the points $2+i$ and $4+3i$). The centre of the circle has to lie on the perpendicular bisector of those two points, and the two points have to be in perpendicular directions from the centre. That requires the centre to be at the point $4+i$ (the green dot), which conveniently shares the same real or imaginary part with each black dot.

Thus $z$ has to lie on the green circle. As CaptainBlack has pointed out, $z$ also has to lie on the red circle centred at $3-i$ with radius 3. At this stage, I would substitute $z=x+iy$, and write the equations of the circles. Solve for $x$ and $y$ (start by subtracting one circle equation from the other, to get a linear relation between $x$ and $y$) and you will get two solutions for $z$, namely the points where the two circles meet.

View attachment 459​

 

[Mr Fantastic]

Notice first that $\arg\left( \dfrac{3z-6-3i}{2z-8-6i}\right) = \arg\left( \dfrac32\,\dfrac{z-2-i}{z-4-3i}\right) = \arg\left( \dfrac{z-2-i}{z-4-3i}\right)$ (because multiplying a complex number by a positive real number does not alter its arg).

Next, $\arg\left( \dfrac{z-2-i}{z-4-3i}\right) = \arg\bigl(z-(2+i)\bigr) - \arg\bigl(z-(4+3i)\bigr)$. Now $z-(2+i)$ is the vector from $z$ to $2+i$, and $z-(4+3i)$ is the vector from $z$ to $4+3i$ (see the dashed lines in the picture below). We want the angle between those two vectors to be $\pi/4$.

If you recall your euclidean geometry (theorems about angles in the same segment, and the angle at the centre being twice the angle at the circumference), you will see that this requires $z$ to lie on a circle through the two black dots in the picture (the points $2+i$ and $4+3i$). The centre of the circle has to lie on the perpendicular bisector of those two points, and the two points have to be in perpendicular directions from the centre. That requires the centre to be at the point $4+i$ (the green dot), which conveniently shares the same real or imaginary part with each black dot.

Thus $z$ has to lie on the green circle. As CaptainBlack has pointed out, $z$ also has to lie on the red circle centred at $3-i$ with radius 3. At this stage, I would substitute $z=x+iy$, and write the equations of the circles. Solve for $x$ and $y$ (start by subtracting one circle equation from the other, to get a linear relation between $x$ and $y$) and you will get two solutions for $z$, namely the points where the two circles meet.

https://www.physicsforums.com/attachments/459​

I think it will be the major arc joining but not including z = 2 + i and z = 4 + 3i rather than the whole circle (the minor arc will be obtained for pi/4 + pi). The end points of the arc are not included because arg(0) is not defined.

 

[Opalg]

I think it will be the major arc joining but not including z = 2 + i and z = 4 + 3i rather than the whole circle (the minor arc will be obtained for pi/4 + pi). The end points of the arc are not included because arg(0) is not defined.

Absolutely correct. (I thought that I had checked what happens on the minor arc, but I overlooked a minus sign.)