find-the-key-value-Arg-z-for-2-i-1-i-

Question Number 130104 by mohammad17 last updated on 22/Jan/21

f i n d t h e k e y v a l u e [A r g (z)] f o r {(2 + i)}^{1 - i}

Answered by Olaf last updated on 22/Jan/21

z = {(2 + i)}^{1 - i}

Z = \ln z = (1 - i) \ln (2 + i)

Z = (1 - i) \ln (\sqrt{5} e^{i \arctan \frac{1}{2}})

Z = (1 - i) [\frac{1}{2} \ln 5 + i \arctan \frac{1}{2}]

Z = (\frac{1}{2} \ln 5 + \arctan \frac{1}{2}) - i (\frac{1}{2} \ln 5 - \arctan \frac{1}{2})

z = e^{Z} = \sqrt{5} e^{\arctan \frac{1}{2}} e^{i (\arctan \frac{1}{2} - \frac{1}{2} \ln 5)}

Ar g z = \arctan \frac{1}{2} - \frac{1}{2} \ln 5 (+ 2 k π)

(or Ar g z = \frac{π - \ln 5}{2} - \arctan 2) (+ 2 k π)

Commented by mohammad17 last updated on 22/Jan/21

t h a n k y o u s i r b a t h o w b e c a m e l n (2 + i) = l n (\sqrt{5} e^{i a r c t a n \frac{1}{2}})

Commented by Olaf last updated on 22/Jan/21

a + i b can be written ρ e^{i θ}

ρ = ∣ a + i b ∣ = \sqrt{a^{2} + b^{2}}

\tan θ = \frac{b}{a}

If a + i b = 2 + i :

ρ = ∣ a + i b ∣ = \sqrt{2^{2} + 1} = \sqrt{5}

\tan θ = \frac{b}{a} = \frac{1}{2} \Rightarrow θ = \arctan \frac{1}{2}

Commented by mohammad17 last updated on 22/Jan/21

t h a n k y o u s i r