Let $z = x + iy$ . We have $z^* = x - iy$ , $iz^* = y + ix$ , $z z^* = x^2 + y^2$ .

Then, $3 + 0i = x+iy + \frac{ 3x - 3y i -y - xi } { x^2 + y^2 } = z + \frac{ 3z^* - i z^* } { z z^*} = z + \frac{ 3 - i } { z}$ .

Clearly, $z =0$ is not a solution, so we can multiply throughout by $z$ to obtain $z^2 - (3) z + (3-i) = 0$ .
Factoring this, we get $\left[ z - (1-i ) \right] \left[ z - (2+i) \right] = 0$ .

Thus, the only possible solutions are $(x,y) = (1, -1), (2, 1 )$ .
We can verify that these are indeed solutions to the original equations.

After looking at the solutions above and the ordered pairs they ended up with, I am having doubts about mine. Could someone please explain where I am going wrong?

$x+\frac { 3x-y }{ { x }^{ 2 }+{ y }^{ 2 } } =2\quad and\quad y-\frac { x+3y }{ { x }^{ 2 }+{ y }^{ 2 } } =0$ Let $z=x+iy$ Multiplying the second equation with $i$ and adding, we get

$z+\frac { (3-i)\bar { z } }{ { \left| z \right| }^{ 2 } } =3$

$=>$ ${ z }^{ 2 }+-3z+(3-i)=0$

$=>$ $z=\quad \frac { 3\pm \sqrt { 9-4(3-i) } }{ 2 }$

And this gives- $x=\frac { 3 }{ 2 } ,iy=\pm \frac { \sqrt { 9-4(3-i) } }{ 2 }$

Here I took a leap of faith; without actually calculating $y$ , I assumed that probably in adding all the solutions $(x,y)$ they will cancel each other and I will end up adding $3/2$ twice. Hence the answer is 3. Now, I am a little skeptic as to whether this is just a coincidence. Could someone please help? Thanks in advance.

(2,1) and (1,-1) satisfies the above pair of equation . Put x=åcos(m) y=åsin(m) After solving both equations in a and m we get å^2=2 or 5 (real values) now we plugin the value of a and get m.

And ordered pair as {(1,1),(1,-1),(-1,1),(-1,-1)} for å^2 as 2 and

{(2,1)(2,-1),(-2,1)(-2,-1)} for å^2=5.

Now check which solutions are added because of squaring while solving for å by putting the values.