Given that Z and H are complex number. obtain the real and imaginary of Z^H


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Question Number 7748 by Tawakalitu. last updated on 13/Sep/16

$G i v e n t h a t Z a n d H a r e c o m p l e x n u m b e r .$ $o b t a i n t h e r e a l a n d i m a g i n a r y o f Z^{H}$
	Answered by Yozzia last updated on 13/Sep/16
	$Let Z=re^(iθ) , H=c+di (r,θ,c,d∈R, r>0, i=(√(−1))). Z^H =(re^(iθ) )^(c+di) =r^(c+di) e^(iθ(c+di)) Z^H =e^((c+di)lnr) e^(cθi−dθ) {∵ r=e^(lnr) } Z^H =e^(clnr) e^(idlnr) e^(−θd) e^(ciθ) Z^H =e^(clnr−θd) e^(idlnr+ciθ) Z^H =e^(clnr−θd) e^(i(dlnr+cθ)) By Euler′s formula, Z^H =e^(clnr−θd) {cos(dlnr+cθ)+isin(dlnr+cθ)} ⇒Re(Z^H )=e^(clnr−dθ) cos(dlnr+cθ) & Im(Z^H )=e^(clnr−dθ) sin(dlnr+cθ) θ=argument of Z, r=modulus of Z. Also, arg(Z^H )=dlnr+cθ & ∣Z^H ∣=e^(clnr−dθ) =e^(−dθ) e^(lnr^c ) =r^c e^(−dθ)$
	$L e t Z = {r e}^{i θ}, H = c + d i (r, θ, c, d \in R, r > 0, i = \sqrt{- 1}) .$ $Z^{H} = {({r e}^{i θ})}^{c + d i} = r^{c + d i} e^{i θ (c + d i)}$ $Z^{H} = e^{(c + d i) l n r} e^{c θ i - d θ} {∵ r = e^{l n r}}$ $Z^{H} = e^{c l n r} e^{i d l n r} e^{- θ d} e^{c i θ}$ $Z^{H} = e^{c l n r - θ d} e^{i d l n r + c i θ}$ $Z^{H} = e^{c l n r - θ d} e^{i (d l n r + c θ)}$ $B y {E u l e r}^{'} s f o r m u l a,$ $Z^{H} = e^{c l n r - θ d} {c o s (d l n r + c θ) + i s i n (d l n r + c θ)}$ $\Rightarrow R e (Z^{H}) = e^{c l n r - d θ} c o s (d l n r + c θ)$ $& I m (Z^{H}) = e^{c l n r - d θ} s i n (d l n r + c θ)$ $θ = a r g u m e n t o f Z, r = m o d u l u s o f Z .$ $A l s o, a r g (Z^{H}) = d l n r + c θ$ $& ∣ Z^{H} ∣= e^{c l n r - d θ} = e^{- d θ} e^{{l n r}^{c}} = r^{c} e^{- d θ}$
	Commented by Tawakalitu. last updated on 13/Sep/16

	$W o w, i r e a l l y a p p r e c i a t e s i r . t h a n k y o u s i r .$
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