Question

If $(a+ib{)}^5=\alpha +i\beta$ then $(b+ia{)}^5$ is equal to

• A. $\beta +i\alpha$
• B. $\alpha -i\beta$
• C. $\beta -i\alpha$
• D. $-\alpha -i\beta$

Answer 1

The correct option is A $\beta +i\alpha$

Let take $a=\cos p$ and $b=\sin p$

$(\cos p+i\sin p)=e^{ip}$

$(\cos p+i\sin p{)}^5=e^{i5p}$

$(\cos 5p+i\sin 5p)=e^{i5p}$

$\cos 5p+i\sin 5p=\alpha +\beta$

we have to calculate

$(b+ai{)}^5=cos(\frac{\pi }{2}-p)+isin(\frac{\pi }{2}-p)=sin5p+icos5p=\beta +i\alpha$