\(\frac{x}{1+i}+\frac{y}{2-i}=\frac{1-5i}{3-2i}\)
⇒ \(\frac{x}{1+i}\times\frac{1-i}{1-i}+\frac{y}{2-i}\times\frac{2+i}{2+i}\) = \(\frac{1-5i}{3-2i}\times\frac{3+2i}{3+2i}\)
⇒\(\frac{x-xi}{1-i^2}+\frac{2y+yi}{4-i^2}\) = \(\frac{3-15i+2i-10i^2}{9-4i^2}\)
(∵ (a + b)(a - b) = a2 - b2)
⇒\(\frac{1}{2}\) (x - xi) + \(\frac{1}{5}\)(2y + yi) = \(\frac{1}{13}\) (13 - 13i) (∵i2 = -i)
⇒ \(\Big(\frac{x}{2}+\frac{2y}{5}\Big)\) + i\(\Big(\frac{-x}{2}+\frac{y}{5}\Big)\) = 1 - i
⇒ \(\frac{x}{2}+\frac{2y}{5}\) = 1 and \(\frac{-x}{2}+\frac{y}{5}\) = -1 (By comparing real & imaginary part of complex numbers)
⇒ \(\frac{5y}{5}\) = 1 - 1 = 0 (By adding both equations)
⇒ y = 0
Put y = 0 in \(\frac{x}{2}+\frac{2y}{5}\)= 1, we get
\(\frac{x}{2}\) = 1
⇒ x = 2
Hence, solution of given complex equation is x = 2, y = 0
