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# Find the real values of $\mathit{x}$ and $\mathit{y}$ for which the following equation: $\left ( 1-\mathit{i} \right )\mathit{x} + \left ( 1+\mathit{i} \right )\mathit{y} = 1-3\mathit{i}$

This is the first part of the multi-part question Q4.
Can you answer this question?

Toolbox:
• If $a+ib=c+id$ then $a=c$ and $b=d$
• (i.e) if $z_1=z_2$,then $Re(z_1)=Re(z_2),Im(z_1)=Im(z_2)$
Step 1:
$(1-i)x+(1+i)y=1-3i$
Separating the real and imaginary parts of LHS
$(x+y)+i(-x+y)=1+3i$
Equating the real and imaginary parts,
$\;x\;+\;\;y=1$------(1)
$-x+\;y=-3$-----(2)
Step 2:
On adding the above two equations we get
$2y=-2$
$y=-1$
Substitute the value of $y$ in equ(1) we get
$x-1=1$
$x=1+1$
$x=2$
$(x,y)=(2,-1)$
answered Jun 7, 2013