# 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.

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.