Answer
Ans.
a. $x=-7/3, y=-24$
b. $x=2/5, y=-1/5$
c. $x=-1, y=0$
Work Step by Step
Given:
a.
$ (3+4i)^2-2(x-iy) =x+iy$
$ (9+24i+16i^2) -2x+2iy=x+iy$
$ 9+24i+16(-1)=(2x + x)+ (-2iy + iy)$
$ (9-16)+24i = 3x- iy$
$-7+24i=3x-iy$
By comparing corresponding real and pure imaginary parts of the LHS and RHS,
$-7 = 3x$ and $24i = -yi$.
Thus,
$x=-7/3 ,y=-24$
Ans.
b.
$ (\frac{1+i}{1-i})^2 + \frac{1}{x+iy} =1+i$
$ (\frac{1+i}{1-i}×\frac{1+i}{1+i})^2+\frac{1}{x+iy} =1+i$
$ \left(\frac{(1+i)^2}{(1-i)(1+i)}\right)^2+\frac{1}{x+iy} =1+i$
$ \left(\frac{1+2i+i^2}{1-i^2}\right)^2+\frac{1}{x+iy} =1+i$
$ \left(\frac{1+2i+(-1)}{1-(-1)}\right)^2+\frac{1}{x+iy} =1+i$
$\left(\frac{2i}{2}\right)^2+\frac{1}{x+iy} =1+i$
$i^2+\frac{1}{x+iy} =1+i$
$-1+\frac{1}{x+iy} =1+i$
$\frac{1}{x+iy} =2+i$
$x+iy =\frac{1}{2+i}$
$x+iy =\frac{1}{2+i}×\frac{2-i}{2-i}$
$x+iy=\frac{2-i}{2^2-i^2}$
$x+iy=\frac{2-i}{4-(-1)}$
$x+iy =\frac{2-i}{5}$
$x+iy=\frac{2}{5}+\left(-\frac{1}{5}\right)i$
By comparing corresponding real and pure imaginary parts of the LHS and RHS,
$x=2/5, y=-1/5$
c.
$(3-2i)(x+iy)=2(x-2yi)+(2i-1) $
$3x+3yi-2xi-2yi^2=2x-4yi+2i-1$
$3x+3yi-2xi-2y(-1)=2x-4yi+2i-1$
$3x+3yi-2xi+2y=2x-4yi+2i-1$
$(3x+2y)+(-2x+3y)i=(2x-1) +(2-4y)i$
On comparing LHS and RHS and equating corresponding real and pure imaginary parts,
$3x+2y=2x-1,\ -2x+3y=2-4y$
$x+2y=-1,\ -2x+7y=2$
After solving the equations above simultaneously (as a linear system of two equations in two unknowns, $x$ and $y$),
$x=-1 , y=0$
Ans.
