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Home  >>  TN XII Math  >>  Complex Numbers
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For what values of $\mathit{x}$ and $\mathit{y}$, the numbers $-3+\mathit{ix^{2}y}$ and $\mathit{x^{2}}+y+4\mathit{i}$ are complex conjugate of each other$?$

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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 $Re(z_1)=Re(z_2),Im(z_1)=Im(z_2)$
Step 1:
$-3+ix^2y$ is the conjugate of $x^2+y+4i$
Therefore $-3-ix^2y=x^2+y+4i$
Equating the real and imaginary parts separately
$x^2+y=-3$-----(1)
$4=-x^2y$------(2)
$\Rightarrow y=\large\frac{-4}{x^2}$-----(3)
Step 2:
Substituting (3) in (1)
$x^2-\large\frac{4}{x^2}$$=-3$
$x^2+3x^2-4=0$
$(x^2+4)(x^2-1)=0$
$x^2=-4,1$ or $x=\pm 2i,\pm 1$
When $x=\pm 2i,y=-1$
When $x=\pm 1,y=-4$
$(x,y)=(\pm 2i,-1),(\pm 1,-4)$
answered Jun 7, 2013 by sreemathi.v
 

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