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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

$${Given}\:{that}\:{Z}\:{and}\:{H}\:{are}\:{complex}\:{number}.\: \\ $$$${obtain}\:{the}\:{real}\:{and}\:{imaginary}\:{of}\:{Z}^{{H}} \\ $$

Answered by Yozzia last updated on 13/Sep/16

Let Z=re^(iθ) , H=c+di  (r,θ,c,d∈R, r>0, i=(√(−1))).  Z^H =(re^(iθ) )^(c+di) =r^(c+di) e^(iθ(c+di))   Z^H =e^((c+di)lnr) e^(cθi−dθ)               {∵ r=e^(lnr) }  Z^H =e^(clnr) e^(idlnr) e^(−θd) e^(ciθ)   Z^H =e^(clnr−θd) e^(idlnr+ciθ)   Z^H =e^(clnr−θd) e^(i(dlnr+cθ))   By Euler′s formula,  Z^H =e^(clnr−θd) {cos(dlnr+cθ)+isin(dlnr+cθ)}  ⇒Re(Z^H )=e^(clnr−dθ) cos(dlnr+cθ)  & Im(Z^H )=e^(clnr−dθ) sin(dlnr+cθ)  θ=argument of Z, r=modulus of Z.  Also, arg(Z^H )=dlnr+cθ   & ∣Z^H ∣=e^(clnr−dθ) =e^(−dθ) e^(lnr^c ) =r^c e^(−dθ)

$${Let}\:{Z}={re}^{{i}\theta} ,\:{H}={c}+{di}\:\:\left({r},\theta,{c},{d}\in\mathbb{R},\:{r}>\mathrm{0},\:{i}=\sqrt{−\mathrm{1}}\right). \\ $$$${Z}^{{H}} =\left({re}^{{i}\theta} \right)^{{c}+{di}} ={r}^{{c}+{di}} {e}^{{i}\theta\left({c}+{di}\right)} \\ $$$${Z}^{{H}} ={e}^{\left({c}+{di}\right){lnr}} {e}^{{c}\theta{i}−{d}\theta} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\because\:{r}={e}^{{lnr}} \right\} \\ $$$${Z}^{{H}} ={e}^{{clnr}} {e}^{{idlnr}} {e}^{−\theta{d}} {e}^{{ci}\theta} \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} {e}^{{idlnr}+{ci}\theta} \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} {e}^{{i}\left({dlnr}+{c}\theta\right)} \\ $$$${By}\:{Euler}'{s}\:{formula}, \\ $$$${Z}^{{H}} ={e}^{{clnr}−\theta{d}} \left\{{cos}\left({dlnr}+{c}\theta\right)+{isin}\left({dlnr}+{c}\theta\right)\right\} \\ $$$$\Rightarrow{Re}\left({Z}^{{H}} \right)={e}^{{clnr}−{d}\theta} {cos}\left({dlnr}+{c}\theta\right) \\ $$$$\&\:{Im}\left({Z}^{{H}} \right)={e}^{{clnr}−{d}\theta} {sin}\left({dlnr}+{c}\theta\right) \\ $$$$\theta={argument}\:{of}\:{Z},\:{r}={modulus}\:{of}\:{Z}. \\ $$$${Also},\:{arg}\left({Z}^{{H}} \right)={dlnr}+{c}\theta\: \\ $$$$\&\:\mid{Z}^{{H}} \mid={e}^{{clnr}−{d}\theta} ={e}^{−{d}\theta} {e}^{{lnr}^{{c}} } ={r}^{{c}} {e}^{−{d}\theta} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 13/Sep/16

Wow, i really appreciate sir. thank you sir.

$${Wow},\:{i}\:{really}\:{appreciate}\:{sir}.\:{thank}\:{you}\:{sir}. \\ $$

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