Help on Algebraic manipulation of a complex number

Question

If $\displaystyle \frac{1}{x+iy}+\frac{1}{u+iv}=1$; x,y,u,v being real quantities, express v in terms of x and y.

My Attempt: $\displaystyle \frac{(u+x)+i(y+v)}{(x+iy)(u+iv)}=1$

$\displaystyle (u+x)+i(y+v)=(xu-yv)+i(xv+yu)$

$\displaystyle (u+x)=(xu-yv)$

$\displaystyle (y+v)=(xv+yu)$

How to manipulate them further ?

Answer 1

Hint:

Let $z=x+iy$ and $w=u+iv$. Then the condition $\frac{1}{x+iy}+\frac{1}{u+iv}=1$ is more readily expressed as:

$$\frac{1}{z}+\frac{1}{w}=1.$$

Solving for $w$,

$$\frac{1}{w}=1-\frac{1}{z}=\frac{z-1}{z}\\ \implies w=\frac{z}{z-1}.$$

Now take the imaginary parts of both sides.

Answer 2

On rearrangement, we have $$u(1-x)+v\cdot y=x\ \ \ \ (1)$$ and $$u\cdot y+v(x-1)-y=0\ \ \ \ (2)$$

Solve for $v$