solving basic complex equations

Question

if I am given a complex equation like this:

\begin{equation} \frac{x+iy+2+3i}{2x+2iy-3}=i+2 \end{equation}

How do I go about solving this equation? Is it necessary to get an expression in the form z=x+iy on the left hand side or is there a shortcut?

Answer 1

$$\frac{x+iy+2+3i}{2x+2iy-3}=i+2$$ we will try what others suggested: $z=x+iy$ $$\frac{z+2+3i}{2z-3}=\frac{(2z-3)(i+2)}{2z-3}$$ $$z+2+3i=(2z-3)(i+2)$$ $$z+2+3i=2zi+4z-3i-6$$ $$3z+2zi-8-6i=0$$ $$3z+2zi=8+6i$$ $$3(x+iy)+2(-y+ix)=8+6i$$ so we can obtain simultaneous equations by taking $\Re$ and $\Im$ parts: $$3x-2y=8$$ $$2x+3y=6$$ now solve

Answer 2

Solving the equation means that you must provide explicit values for the unknowns, $x$ and $y$.

Using complex number, you can write

$$\frac{z+2+3i}{2z-3}=2+i$$ and solve for $z$ by symbolic manipulation.

This yields

$$(2(2+i)-1)z=2+3i+3(2+i),$$ then

$$z=\frac{2+3i+3(2+i)}{2(2+i)-1}.$$

Now you do have to extract the real and imaginary parts, as $x,y$ are requested.