find the key value [Arg(z)]for (2+i)^(1−i)


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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

	$t h a n k y o u s i r$
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