Question

If $a,b,c,d,p$ are real such that $(a+ib{)}^{c+id}=p>0$, then $(a^2+b^2{)}^{(c^2+d^2)}=$

• A. $p^c$
• B. $p^{2c}$
• C. $p^d$
• D. $p^{2d}$

Answer 1

The correct option is A $p^c$

${(a+ib)}^{c+id}=p$
$(c+id)\log \left(a+ib\right)=\log p$
$\frac{(c^2+d^2)}{c-id}\log \frac{(a^2+b^2)}{(a-ib)}=\log p$
$(c^2+d^2)\log (a^2+b^2)-(c^2+d^2)\log (a-ib)=(c-id)\log p$
$\Rightarrow \log {(a^2+b^2)}^{c^2+d^2}-(c^2+d^2)\log (a-ib)=c\log p-id\log p$
Comparing both sides with real and img. part
$\log {(a^2+b^2)}^{(c^2+d^2)}=\log p^c$
${(a^2+b^2)}^{(c^2+d^2)}=p^c$