Question

If $a+i{b}^{11}$ = p + iq, where i = $\sqrt{-1}$.Find the value of $(b+ia{)}^{11}$.

• A. -q - ip
• B. q + ip
• C. p + iq
• D. -p - iq

Answer 1

The correct option is A

-q - ip

Given $(a+ib{)}^{11}$ = p + iq

We know, from complex number properties $(\overline{Z}{)}^2$ = ($\frac{}{Z^2}$)

${\overline{(a+ib)}}^{11}$ = $\frac{}{{(a+ib)}^{11}}$

Substitute the value of $(a+ib{)}^{11}$ = p + iq

We get,

$(a-ib{)}^{11}$ = p - iq

Taking $(-i{)}^{11}$ common in LHS and -i common in RHS

$(-i{)}^{11}$ $[b+ia{]}^{11}$ = (-i) (q + ip)

$(-1{)}^{11}$ $[b+ia{]}^{11}$ = (-i) (q + ip)

Substitute the value of $(i{)}^{11}$ = -i

(-1) (-i)$(b+ia{)}^{11}$ = (-i)(q + ip)

$(b+ia{)}^{11}$ = - q - ip

So, the value of $(b+ia{)}^{11}$ = - q - ip