Question

If $z=a+ib$ then its conjugate is $a-ib$. If $1,\omega ,{\omega }^2$ are cube roots of unity then (i) $1+\omega +{\omega }^2=0$ (ii) ${\omega }^3=1$ The conjugate of $\frac{6-3i}{7+i}$ is

• A. $\frac{39-27i}{50}$
• B. $\frac{-39+27i}{50}$
• C. $\frac{39+27i}{50}$
• D. $\frac{-39-27i}{50}$

Answer 1

Answer: (C)

$\frac{39+27i}{50}$

$Z=\frac{6-3i}{7+i}$

$Z=\frac{(6-3i)(7-i)}{(7+i)(7-i)}$

$Z=\frac{42-6i-21i-3}{(7{)}^2-(i{)}^2}$

$Z=\frac{39-27i}{49+1}$

$Z=\frac{39-27i}{50}$

$\overline{Z}=\overline{\frac{39-27i}{50}}$

$\overline{Z}=\frac{39+27i}{50}$

Conjugate of $\frac{6-3i}{7+i}=\frac{39+27i}{50}$