# Complex number question - spurious solutions…

$$z_1 = 2 + 3i$$ 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.

## 3 Answers

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

• Thanks, this is amazing – PhysicsMathsLove Aug 18 '18 at 21:14

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.

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.