# Math Help - real numbers x and y

1. ## real numbers x and y

if someone could help me with this i would appreciate it.

given that

$\frac{1}{x+jy}+\frac{1}{1+2j} =1$

find the real numbers x and y

many thanks

2. Originally Posted by decoy808
if someone could help me with this i would appreciate it.

given that

$\frac{1}{x+jy}+\frac{1}{1+2j} =1$

find the real numbers x and y

many thanks
"Rationalize" the denominators by multiplying numerator and denominator of each fraction by the complex conjugate of the denominator:
$\frac{1}{x+ jy}\frac{x- jy}{x- jy}+ \frac{1}{1+2j}\frac{1-2j}{1-2j}= 1$
$\frac{x- jy}{x^2+y^2}+ \frac{1-2j}{5}= 1$

Get rid of the fractions by multiplying both sides by $5(x^2+y^2)$
$5x- 5jy+ (x^2+y^2)- 2(x^2+y^2)j= 5x^2+5y^2$

Finally, set "real part" equal to "real part" and "imaginary part" equal to imaginary part:
$5x+ x^2+y^2= 5x^2+ 5y^2$
and
$-5y- 2x^2- 2y^2= 0$

Solve those two equations for x and y.

3. Hello, decoy808!

Given: . $\frac{1}{x+yj}+\frac{1}{1+2j} \:=\:1$

find real numbers $x$ and $y.$

Multiply through by the LCD:

. . $(x+yj)(1 + 2j)\cdot\frac{1}{x+yj} + (x+yj)(1+2j)\cdot\frac{1}{1+2j} \;=\;(x+yj)(1+2j)\cdot 1$

. . $(1 + 2j) + (x+yj) \:=\:(x+yj)(1 + 2j)$

. . $1 + 2j + x + yj \:=\:x + 2xj + yj + 2yj^2$

. . $1 + 2j + x + yj \;=\;x + 2xj + yj - 2y$

. . $(1 + x) + (2 + y)j \;=\;(x-2y) + (2x+y)j$

Equate real and imaginary components:

. . $\begin{array}{cccccccccccc}
1 + x \:=\: x-2y & \Rightarrow & 1 \:=\: \text{-}2y & \Rightarrow &\boxed{ y \:=\: \text{-}\tfrac{1}{2}} \\ \\[-3mm]
2+y \L=\: 2x +y & \Rightarrow & 2 \:=\: 2x & \Rightarrow & \boxed{x \:=\: 1} \end{array}$