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

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

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 .