Question

Find the real values of x and y, if

$\left(i\right)(x+iy)(2-3i)=4+i\left(ii\right)(3x-2iy)(2+i{)}^2=10(1+i\left)\right(iii)\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i(iv\left)\right(1+i\left)\right(x+iy)=2-5i$

Answer 1

$\left(i\right)\text{We have}(x+iy)(2-3i)=4+i\Rightarrow x(2-3i)+iy(2-3i)=4+i\Rightarrow 2x-3xi+2yi+3y=4+i\Rightarrow 2x+3y+i(-3x+2y)=4+i$

Equating the real and imaginary parts we get

$2x+3y=4....\left(i\right)-3x+2y=1....\left(ii\right)$

Multiplying (i) by 3 and (ii) by 2 and adding

$6x-6x-9y+4y=12+2\Rightarrow 13y=14\Rightarrow y=\frac{14}{13}$

Substituting the value of y in (i), we get

$2x+3\times \frac{14}{13}=4\Rightarrow 2x+\frac{42}{13}=4\Rightarrow 2x=4-\frac{42}{13}\Rightarrow 2x=\frac{52-42}{13}\Rightarrow 2x=\frac{10}{13}\Rightarrow x=\frac{5}{13}\text{Hence}x=\frac{5}{13}\text{and}y=\frac{14}{13}$

$\left(ii\right)(3x-2iy)(2+i{)}^2=10(1+i)\Rightarrow (3x-2iy\left)\right(2^2+i^2+2\times 2\times i)=10+10i\Rightarrow (3x-2iy\left)\right(4-1+4i)=10+10i\Rightarrow 3x(3+4i)-2iy(3+4i)=10+10i\Rightarrow 9x+12xi-6yi+8y=10+10i\Rightarrow 9x+8y+i(12x-6y)=10+10i$

Equating the real and imaginary parts we get

9x + 8y = 10 ...... (i)

12x - 6y = 10 .......(ii)

Multiplying (i) by 6 and (ii) by 8 and adding

$54x+96x+48y-48y=60+80\Rightarrow 150x=140\Rightarrow x=\frac{140}{150}\Rightarrow x=\frac{14}{15}$

Substituting value of x in (i) we get

$9\times \frac{14}{15}+8y=10\Rightarrow \frac{42}{5}+8y=10\Rightarrow 8y=10-\frac{42}{5}\Rightarrow 8y=\frac{50-42}{5}\Rightarrow 8y=\frac{8}{5}\Rightarrow y=\frac{1}{5}\text{Hence,}x=\frac{14}{5}\text{and}y=\frac{1}{5}$

$\left(iii\right)\frac{(1+i}{x}-2i3+i+\frac{(2-3i)y+i}{3-i}=i\Rightarrow \frac{(3-i)\left(\right(1+i)x-2i)+(3+i)\left(\right(2-ei\left)\right(y+i\left)\right)}{(3+i)(3-i)}=i\Rightarrow \frac{(3-i)(1+i)x-2i(3-i)+(3+i)(2-3i)y+i(3+i)}{3^2+1^2}=i\Rightarrow \frac{(3+3i-i+1)x-6i-2+(6-9i+2i+3)y+3i-1}{9+1}=i\Rightarrow \frac{(4+2i)x-6i-2+(9-7i)y+3i-1}{10}=1\Rightarrow 4x+2ix-6i-2+9y-7iy+3i-1=10i\Rightarrow 4x+9y-3+i(2x-7y-3)=10i$

Equating real and imaginary parts we get

4x + 9y - 3 = 0 .... (i)

and 2x - 7y - 3 = 10

i.e. 2x - 7y = 13

Multiplying (i) by 7, (ii) by 9 and adding we get

$28x+18x+63y-63y=117+21\Rightarrow 46x=117+21\Rightarrow 46x=138\Rightarrow x=\frac{138}{46}=3$

Substituting the value of x = 3 in (i), we get

$4\times 3+9y=3\Rightarrow 9y=-9\Rightarrow y=\frac{-9}{9}\Rightarrow y=-1Hence,x=3,y=-1$

$\left(iv\right)(1+i)(x+iy)=2-5i\Rightarrow 1(x+iy)+i(x+iy)=2-5i\Rightarrow x+iy+ix-y=2-5i\Rightarrow x-y+i(x+y)=2-5i$

Equating real and imaginary parts we get

x - y = 2 .....(i)

x + y = -5 .....(ii)

Adding (i) and (ii) we get

2x = 2 - 5

$\Rightarrow$ 2x = -3

$\Rightarrow x=\frac{-3}{2}$

Substituting the value of x in (i), we get

$\frac{-3}{2}-y=2\Rightarrow \frac{-3}{2}=y\Rightarrow y=\frac{-3-4}{2}\Rightarrow y=\frac{-7}{2}\text{Hence}x=\frac{-3}{2},y=\frac{-7}{2}$