Question

Given that $x,y\in R$ solve:
(a) $(x+2y)+i(2x-3y)=5-4i$
(b) $(x+iy)+(7-5i)=9+4i$
(c) $x^2-y^2-i(2x+y)2i$

Answer 1

In complex equations

Real part is equal to real part and imaginary part is equal to imaginary part

Therefore

(a)$(x+2\times y)+i\times (2\times x-3\times y)=(5-(4\times i\left)\right)$

On equating we get $(x+(2\times y\left)\right)=5$ and$(2\times x-3\times y)=-4$

On solving above equation we get $x=1$ and $y=2$

(b)on rearranging we get

$(x+7)+i\times (y-5)=9+4\times i$

On solving we get $x=2$ and $y=9$

(c) we know that $i\times i=i^2=-1$

Therefore on solving we get

$x\times x-y\times y-2(2\times x+y)(i\times i)$

$x^2-y^2+2(2\times x+y)$

$x^2-y^2+(4\times x+y)$