prove-that-Arg-log-3-4i-1-i-log-3-4i-5-help-me-sir-please-

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