Question

The value of x and y which satisfies the equation :
$\frac{\left(1+i\right)^2}{\left(1-i\right)^2}+\frac{1}{x+iy}=1+i$ is -​

Answer 1

We have from Question :-

$\frac{ {(1 + i)}^{2} }{ {(1 - i)}^{2} } + \frac{1}{x + iy} = 1 + i \\ \\ \\ \\ \frac{1 + {i}^{2} + 2i}{1 + {i}^{2} - 2i} + \frac{1}{x + iy} = 1 + i \\ \\ \\ \\ \frac{1 - 1 + 2i}{1 - 1 - 2i} + \frac{1}{x + iy} = 1 + i \\ \\ \\ \\ - 1 + \frac{1}{x + iy} = 1 + i \\ \\ \\ \\ \frac{1}{x + iy} = 2 + i \\ \\ \\ \implies x + iy = \frac{1}{2 + i} \\ \\ \\ \\x + iy = \frac{2 - i}{(2 + i)(2 - i)} \\ \\ \\x + iy = \frac{2 - i}{4 - {i}^{2} } \\ \\ \\x + iy = \frac{2 - i}{4 - ( - 1)} \\ \\ \\x + iy = \frac{2 - i}{5} \\ \\ \\ x + iy = \frac{2}{5} + \frac{( - 1)}{5}i \\ \\ \\ \\on \: comparing \: both \: sides \: \\ \\ \\x = \frac{2}{5} \\ \\ \\y = \frac{ - 1}{5}$

$\rule{200}{2}$

Hence,

x = 2/5

y = - 1/5

$\rule{200}{2}$

Answer 2

Answer:

Step-by-step explanation:

Complex numbers:-

Given an equation,in which we have to find x,y

Applying (a+b)^2 ,(a-b)^2 formula,we get:-

1/(x+iy)=2+i

Taking reciprocal,and multiplying by 2-i at both numerator and denominator,we get:-

x=2/5,y=-1/5