prove that: Arg(log(3−4i))=(1/i)log(((3−4i)/5)) help me sir please


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Question Number 127364 by mohammad17 last updated on 29/Dec/20

$p r o v e t h a t : A r g (l o g (3 - 4 i)) = \frac{1}{i} l o g (\frac{3 - 4 i}{5})$ $h e l p m e s i r p l e a s e$
Commented by mohammad17 last updated on 29/Dec/20

$p l e a s h e l p m e ?$
	Answered by ebi last updated on 29/Dec/20

	$z = l o g (3 - 4 i) = l o g (a + b i)$ $l e t a + b i = {r e}^{i θ} (c o n v e r t t o p o l a r f o r m)$ $w h e r e r = \sqrt{a^{2} + b^{2}} a n d θ = a r g (a + b i)$ $r = \sqrt{3^{2} + {(- 4)}^{2}} = \sqrt{25} = 5$ $l o g (3 - 4 i) = l o g (5 e^{i θ})$ $l o g (3 - 4 i) = l o g (5) + i θ$ $i θ = l o g (3 - 4 i) - l o g (5)$ $i θ = l o g (\frac{3 - 4 i}{5})$ $θ = \frac{1}{i} l o g (\frac{3 - 4 i}{5})$ $∴ a r g (3 - 4 i) = \frac{1}{i} l o g (\frac{3 - 4 i}{5})$
	Commented by mohammad17 last updated on 29/Dec/20

	$y o u a r e a w o n d e r f u l p e r s o n, t h a n k y o u s i r$
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