complex numbers

**thereddevils** · September 11th 2009, 06:43 AM

Find the real numbers x and y if $\frac{1}{x+iy}+\frac{1}{1+2i}=1$

I started like this :

Multiply their respective conjugates for both terms :

i got $\frac{x-iy}{x^2+y^2}+\frac{1-2i}{5}=1$
continue solving from here ,

$<br /> \frac{x^2+y^2+5x}{5x^2+5y^2}-[\frac{5y+2x^2+2y^2}{5x^2+5y^2}]i=1<br />$

so i ot 2 equations here :

$5x=4x^2+4y^2$

$5y+2x^2+2y^2=0$

I am not sure how to continue from here .. not even sure if my above working is correct . THanks .

**red_dog** · September 11th 2009, 09:14 AM

$\left\{\begin{array}{ll}4x^2+4y^2-5x=0\\2x^2+2y^2+5y=0\end{array}\right.$

Multiply the second ecuation by 2 and substract from the first:

$-5x-10y=0\Rightarrow x=-2y$

Replace x in the second equation:

$10y^2+5y=0$

Then $y_1=0\Rightarrow x_1=0$ which are not good

and $y_2=-\frac{1}{2}\Rightarrow x_2=1$

**ramiee2010** · September 11th 2009, 09:37 AM

it can be done by this way

Thread: complex numbers

LinkBack

Thread Tools

Display

complex numbers

Similar Math Help Forum Discussions

raising complex numbers to complex exponents?

Finding real numbers in equations containing complex numbers

What is complex slope of a line in complex numbers

Pre-Calc Graphing with Complex numbers. Multiplying with complex numbers.

Complex Numbers- Imaginary numbers

Search Tags