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Home  >>  TN XII Math  >>  Complex Numbers
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Find the real values of $\mathit{x}$ and $\mathit{y}$ for which the following equation: $\large\frac{\left ( 1+\mathit{i} \right )\mathit{x}-2\mathit{i}}{3+\mathit{i}} + \frac{\left ( 2-3\mathit{i} \right )\mathit{y}+\mathit{i}}{3-\mathit{i}} = $$\mathit{i}$

This is the second part of the multi-part question Q4.

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1 Answer

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Toolbox:
  • If $a+ib=c+id$ then $a=c$ and $b=d$
  • (i.e) if $z_1=z_2$,then $Re(z_1)=Re(z_2),Im(z_1)=Im(z_2)$
  • If $z_1=a+ib$, $z_2=c+id$
  • $z_1z_2=(a+ib)(c+id)=(ac-bd)+i(ad+bc)$
  • $\mid z_1z_2\mid=\mid z_1\mid\mid z_2\mid$
Step 1:
$\large\frac{(1+i)x-2i}{3+i}+\large\frac{(2-3i)y+i}{3-i}$$=i$
Simplifying the terms on the LHS
$\large\frac{(1+i)x-2i}{3+i}=\frac{x+i(x-2)}{3+i}$
$\qquad\qquad\;\;\;=\large\frac{[x+i(x-2)][3-i]}{(3+i)(3-i)}$
$\qquad\qquad\;\;\;=\large\frac{3x-(x-2)(-1)+[3(x-2)-x]i}{9+1}$
$\qquad\qquad\;\;\;=\large\frac{(4x-2)+i(2x-6)}{10}$-----(1)
$\large\frac{(2-3i)y+i}{3-i}=\frac{2y+i(-3y+1)}{3-i}$
$\qquad\qquad\;\;=\big(\large\frac{(2y+i(1-3y))}{3-i}\big)\big(\large\frac{3+i}{3+i}\big)$
$\qquad\qquad\;\;=\large\frac{6y-(1-3y)+i(2y+3-9y)}{9+1}$
$\qquad\qquad\;\;=\large\frac{(9y-1)+i(-7y+3)}{10}$-----(2)
$(1)+(2)$ we get
$\Rightarrow LHS= \large\frac{(4x+9y-3)+i(2x-7y-3)}{10}$$=i(RHS)$
Step 2:
Equating the real and imaginary parts and simplifying
$4x+9y=3$
$2x-7y=13$
$\Delta=\begin{vmatrix}4 & 9\\2 & -7\end{vmatrix}$
$\quad=-28-18=-46$
$\Delta x=\begin{vmatrix}3& 9\\13 & -7\end{vmatrix}$
$\quad\;=-21-117=-138$
$\Delta x=-138$----(3)
$\Delta y=\begin{vmatrix}4& 3\\2 & 13\end{vmatrix}$
$\quad\;=52-6=46$
$\Delta y=46$----(4)
Step 3:
Substitute the value of $\Delta$ in eq(3) we get
$\Delta x=-138$
$x=\large\frac{-138}{-46}$$=3$
Substitute the value of $\Delta$ in eq(4) we get
$\Delta y=46$
$y=\large\frac{46}{-46}$$=-1$
$(x,y)=(3,-1)$
answered Jun 7, 2013 by sreemathi.v
edited Jun 7, 2013 by sreemathi.v
 

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