Question Number 7748 by Tawakalitu. last updated on 13/Sep/16
$${Given}\:{that}\:{Z}\:{and}\:{H}\:{are}\:{complex}\:{number}.\: \\ $$$${obtain}\:{the}\:{real}\:{and}\:{imaginary}\:{of}\:{Z}^{{H}} \\ $$
Answered by Yozzia last updated on 13/Sep/16
$${Let}\:{Z}={re}^{{i}\theta} ,\:{H}={c}+{di}\:\:\left({r},\theta,{c},{d}\in\mathbb{R},\:{r}>\mathrm{0},\:{i}=\sqrt{−\mathrm{1}}\right). \\ $$$${Z}^{{H}} =\left({re}^{{i}\theta} \right)^{{c}+{di}} ={r}^{{c}+{di}} {e}^{{i}\theta\left({c}+{di}\right)} \\ $$$${Z}^{{H}} ={e}^{\left({c}+{di}\right){lnr}} {e}^{{c}\theta{i}−{d}\theta} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\because\:{r}={e}^{{lnr}} \right\} \\ $$$${Z}^{{H}} ={e}^{{clnr}} {e}^{{idlnr}} {e}^{−\theta{d}} {e}^{{ci}\theta} \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} {e}^{{idlnr}+{ci}\theta} \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} {e}^{{i}\left({dlnr}+{c}\theta\right)} \\ $$$${By}\:{Euler}'{s}\:{formula}, \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} \left\{{cos}\left({dlnr}+{c}\theta\right)+{isin}\left({dlnr}+{c}\theta\right)\right\} \\ $$$$\Rightarrow{Re}\left({Z}^{{H}} \right)={e}^{{clnr}−{d}\theta} {cos}\left({dlnr}+{c}\theta\right) \\ $$$$\&\:{Im}\left({Z}^{{H}} \right)={e}^{{clnr}−{d}\theta} {sin}\left({dlnr}+{c}\theta\right) \\ $$$$\theta={argument}\:{of}\:{Z},\:{r}={modulus}\:{of}\:{Z}. \\ $$$${Also},\:{arg}\left({Z}^{{H}} \right)={dlnr}+{c}\theta\: \\ $$$$\&\:\mid{Z}^{{H}} \mid={e}^{{clnr}−{d}\theta} ={e}^{−{d}\theta} {e}^{{lnr}^{{c}} } ={r}^{{c}} {e}^{−{d}\theta} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 13/Sep/16
$${Wow},\:{i}\:{really}\:{appreciate}\:{sir}.\:{thank}\:{you}\:{sir}. \\ $$