Solve for $x$ and $y$ in $\left( \frac{1+i}{1-i} \right)^2+\frac{1}{x+yi}=2+5i$

Question

$ \left (\frac{1+i}{1-i} \right)^2+\frac{1}{x+y i}=2 + 5 i$

I can't figure out where to start.

Answer 1

Hint:

Is obvious that $x,y$ they are not both zero.

Observe: $(\frac{1+i}{1-i})^{2} + \frac{1}{x+iy} = 2 + 5i$ $$\frac{1}{x+iy}=2 + 5i-((\frac{1+i}{1-i})^{2})$$ $$\frac{1}{x+iy}=2 + 5i-(-1)$$ $$\frac{1}{x+iy}=3 + 5i$$ $$\frac{x-iy}{x^{2}+y^{2}}=3 + 5i$$

Then, $$\frac{x}{x^{2}+y^{2}}=3$$$$\frac{-y}{x^{2}+y^{2}}=5$$

So, $$\frac{x}{3}=x^{2}+y^{2}=\frac{-y}{5}$$ thus $$x=\frac{-3y}{5}$$ and replacing above is easily obtained $$y=\frac{-5}{34}, x= \frac{3}{34}$$

Answer 2

Isolate 1/(x+yi), by subtracting the rest to the right side. Find a common denominator on the right and invert both sides. Now x+iy has been isolated and x=Re(RHS) and y=Im(RHS) (where RHS stands for right hand side).

Alternatively, instead of finding a common denominator on the right hand side you can simplify the portion with $(1+i)/(1-i)$.

Answer 3

HINT

\begin{align*} \begin{cases} \displaystyle\frac{1+i}{1-i} = \frac{1+i}{1-i}\times\frac{1+i}{1+i} = \frac{(1+i)^{2}}{1-i^{2}} = \frac{2i}{2} = i\\\\ \displaystyle\frac{1}{x+yi} = \frac{1}{x+yi}\times\frac{x-yi}{x-yi} = \frac{x-yi}{x^{2}+y^{2}} \end{cases} \end{align*}

Answer 4

Here's a way to reduce the amount of computation. $$ {\left( {\frac{{1 + i}}{{1 - i}}} \right)^2} = - 1 $$ $$ \frac{1}{{x + iy}} = 3 + 5i $$ $$ \left( {3 + 5i} \right)\left( {x + iy} \right) = 1 \to \left( {3x - 5y} \right) + i\left( {5x + 3y} \right) = 1 $$ Then,

$$ \begin{cases} 3x-5y=1\\ 5x+3y=0 \end{cases} $$

and the answer will be easily obtained.