Question:- Find all the complex numbers $z$ for which $\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4}$ and $|z-3+i|=3$

My solution:-

$\begin{equation} \arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4} \implies \arg\left(\dfrac{z-2-i}{z-4-3i}\right)=\dfrac{\pi}{4} \end{equation}\tag{1}$

Now, what this suggests is that $z$ lies on a segment with the end points of the segment being $(2,1)$ and $(4,3)$. Now to find the equation of the circle we find the center and radius of the circle which comes out to be $(4,1)$ and $2$, respectively. So, the equation of the circle comes out to be


Now all we need to do is find the point of intersection of the circles $|z-4-i|=2$ and $|z-3+i|=3$ to find the desired $z$, which comes out to be

$$\boxed {z=\left(4 \mp \dfrac{4}{\sqrt{5}}\right) + i\left(1 \pm \dfrac{2}{\sqrt{5}} \right)}$$.

Now just to check which one is the solution I substituted both these $z$ back into $(1)$ but I found out that neither satisfy the condition as the argument comes out to be $\dfrac{\pm\sqrt{5}+5}{2}$, instead of 1, so according to my solution neither are the answer.

My problem with the question:- But as it turns out as per the book I am solving, the answers are both the complex numbers $z$ that I found out. Also, I think only one of these is the solution because on some rough plotting I found out that the circle $|z-3+i|=3$ intersects the segment $|z-4-i|=2$ with end points as mentioned in the solution only once at $z=\left(4+\dfrac{4}{\sqrt{5}}\right) + i\left(1-\dfrac{2}{\sqrt{5}} \right)$.

So, is it me that is wrong or the book or maybe both. If there is anything wrong in my solution please point it out.

  • $\begingroup$ The first locus expression is not a whole circle but a part-circle $\endgroup$ Jul 11, 2016 at 15:38
  • $\begingroup$ As you see in my solution I have indeed considered it as a segment, stating its end points too. $\endgroup$
    – user350331
    Jul 11, 2016 at 15:39
  • $\begingroup$ Try writing $z = 3-i + 3e^{i\theta}$ $\endgroup$
    – C Marius
    Jul 11, 2016 at 15:50
  • $\begingroup$ Clearly there should be only one solution, as your diagram shows $\endgroup$ Jul 11, 2016 at 16:02
  • $\begingroup$ @DavidQuinn:- So what should be the answer to this one according to you. $\endgroup$
    – user350331
    Jul 11, 2016 at 16:13

2 Answers 2


From the following where $\circ=\pi/4$,

$\qquad\qquad\qquad $enter image description here

the circle $$(x-4)^2+(y-1)^2=4$$ you've found is correct, but note that we have only the part below the line $y=x-1$ passing through $(2,1)$ and $(4,3)$.

This is because the angle (as measured in counter-clock wise direction) from $\vec{zz_1}$ to $\vec{zz_2}$ is $\pi/4$ where $z_1=4+3i,z_2=2+i$.

$\qquad\qquad\qquad$enter image description here

Thus, solving $$(x-4)^2+(y-1)^2=4$$ $$y\lt x-1$$ $$(x-3)^2+(y+1)^2=3^2$$ gives only one solution $$x=4+\frac{4}{\sqrt 5},\quad y=1-\frac{2}{\sqrt 5}$$ (and this satisfies $(1)$.)




So, we get when $z=\Re[z]+\Im[z]i$:

$$\text{Q}=\frac{3z-6-3i}{2z-8-6i}=\frac{3\Re[z]-6+3i\left(\Im[z]-1\right)}{2\Re[z]-8+2i\left(\Im[z]-3\right)}=$$ $$\frac{\left(3\Re[z]-6+3i\left(\Im[z]-1\right)\right)\left(2\Re[z]-8-2i\left(\Im[z]-3\right)\right)}{\left(2\Re[z]-8+2i\left(\Im[z]-3\right)\right)\left(2\Re[z]-8-2i\left(\Im[z]-3\right)\right)}=$$ $$\frac{6\Re^2[z]-36\Re[z]+6\Im^2[z]-24\Im[z]+66+12i\left(\Re[z]-\Im[z]-1\right)}{\left(2\Re[z]-8\right)^2+\left(2\Im[z]-6\right)^2}$$

So, when $\Re[\text{Q}]=\Im[\text{Q}]$ we get:

$$6\Re^2[z]-36\Re[z]+6\Im^2[z]-24\Im[z]+66=12\left(\Re[z]-\Im[z]-1\right)\Longleftrightarrow$$ $$\Re^2[z]-6\Re[z]+\Im^2[z]-4\Im[z]+11=2\left(\Re[z]-\Im[z]-1\right)\Longleftrightarrow$$ $$6(\Re[z]-4)^2+6(\Im[z]-1)^2-24=0\Longleftrightarrow$$ $$(\Re[z]-4)^2+(\Im[z]-1)^2=4$$

Now, given is that $|z-3+i|=3$, so:


So, you need to solve:

$$ \begin{cases} (\Re[z]-4)^2+(\Im[z]-1)^2=4\\ \left(\Re[z]-3\right)^2+\left(1+\Im[z]\right)^2=9 \end{cases} $$

When you solve the system, you get two solutions:

  1. $$\Re[z]=4-\frac{4}{\sqrt{5}},\Im[z]=1+\frac{2}{\sqrt{5}}$$
  2. $$\Re[z]=4+\frac{4}{\sqrt{5}},\Im[z]=1-\frac{2}{\sqrt{5}}$$

But, the first one does not work because of the argument.

So, the only solution is:



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