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If $\large\frac{c+i}{c-i}$$=a+ib$, then $a^2+b^2$ = ?

(A) 1
(B) $1+c^2$
(C) $c^2$
(D) $1-c^2$

1 Answer

Toolbox:
  • $z\overline z=|z|^2$
  • $\large(\overline{\frac{z_1}{z_2}})=\frac{\overline z_1}{\overline z_2}$
Given: $a+ib=\large\frac{c+i}{c-i}$
$\Rightarrow\:\overline {a+ib}=a-ib=\large\frac{c-i}{c+i}$
$(a+ib)(a-ib)=a^2+b^2=\large\frac{c+i}{c-i}\times\frac{c-i}{c+i}$$=1$
answered Jul 16, 2013 by rvidyagovindarajan_1
 

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