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$z_1 = 2 + 3i$ and $z_2 = 3 - 4i$

The complex number $z = x + iy$ is such that $\frac{z + z_1}{2z - z_2} = 1$.

Find the value of $x$ and the value of $y$.

Method 1:

$$z + z_1 = 2z - z_2 $$$$\Rightarrow x + iy + 2 + 3i = 2x + 2iy - 3 + 4i$$

Equating real and imaginary components, I obtain $x = 5$ and $y = -1$.

This is a single unique solution as expected.

Method 2:

$$\frac{x + iy + 2 + 3i}{2x + 2iy - 3 + 4i} = 1$$

$$\Rightarrow \frac{(x+2) + i(y+3)}{(2x-3) + i(2y + 4)} = 1$$ Multiplying top and bottom by the conjugate of the denominator gives: $$\frac{(x+2)(2x-3) + (y+3)(2y+4)}{(2x-3)^2 + (2y+4)^2} + i \frac{(y+3)(2x-3) - (x+2)(2y+4)}{(2x-3)^2 + (2y+4)^2} = 1$$

Straightaway, I can see that this will have two solutions for $x$ and $y$ when you equate components as one of the equations will be quadratic.

Working through the algebra gives $x= 5$ and $y=-1$ as above but also $x = \frac{3}{2}$ and $y = -2$. I have noticed that this additional solution gives a denominator of $0$ when substituted back into the problem, so I suppose the issue lies there somehow...


Question: I would like to be enlightened about the origins of this additional solution and why it appears in this 2nd method and not the 1st.

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As you noticed the extra solution is not acceptable since it is a zero of the denominator. In the general setting if you have a fraction of the same kind $\displaystyle \frac{f(z)}{g(z)} = 0$ then mupliplyng and dividing by the complex coniugate of $g$ yields $$ \frac{f(z) \overline{g(z)}}{|g(z)|} = 0 $$ Hence it adds to the numerator all the zeroes of $\overline{g(z)}$ but it is always true that $$ \overline{g(z)} = 0 \quad \iff \quad |g(z)| = 0 \quad \iff \quad g(z) = 0 $$ Hence you are not adding new solutions

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    $\begingroup$ Thanks, this is amazing $\endgroup$ – PhysicsMathsLove Aug 18 '18 at 21:14
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When you write a fraction you always have to justify why the denominator is not null. So, when you're trying to solve the equation :

$$ \frac{z_1 + z}{2z - z_2} = 1$$

You have to add : "for $2z \neq z_2$" i.e. for $x \neq 3/2$ and $y\neq -2$. So those numbers aren't a solution.

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We have that

$$\frac{a}{b}=1\implies \frac{ab}{b^2}=1\implies ab=b^2\implies b=a\:\text{or}\: b=0.$$ But $b=0$ is not a solution of $\frac{a}{b}=1.$

For the same reason

$$\dfrac{1}{z}=2\implies \dfrac{\bar{z}}{z\bar{z}}=2\implies \bar{z}=2|z|^2.$$ $z=0$ is a solution of $\bar{z}=2|z|^2$ but not of $\dfrac{1}{z}=2.$

Conclusion: We have to be careful that we don't multiply and divide by a quantity that can be zero.

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