S-log-a-i-i-2-1-a-R-Solve-S-extra-what-if-a-C-

Question Number 4582 by FilupSmith last updated on 08/Feb/16

S = \log_{a} i

i^{2} = - 1

a \in R

S o l v e S

e x t r a :

what if a \in C ?

Answered by Yozzii last updated on 09/Feb/16

B y c h a n g e o f b a s e,

S = {l o g}_{a} i = \frac{l n i}{l n a} .

N o w, i = 0 + i \times 1 = c o s α + i s i n α

∴ s i n α = 1 \Rightarrow α = n π + {(- 1)}^{n} \frac{π}{2} (n \in Z)

c o s α = c o s (n π + {(- 1)}^{n} 0.5 π) = {(- 1)}^{n} \times 0 + 0 \times 1 = 0

∴ i = e^{i (n π + {(- 1)}^{n} \frac{π}{2})}

\Rightarrow S = \frac{l n e x p (i {n π + \frac{{(- 1)}^{n} π}{2}})}{l n a}

S = \frac{i}{l n a} (n π + \frac{π {(- 1)}^{n}}{2})

(1) I f a > 0, t h e n l n a \in R a n d i n d e e d S = \frac{i}{l n a} (n π + \frac{π {(- 1)}^{n}}{2}) (n \in Z) .

S o, S c a n t a k e i n f i n i t e l y m a n y v a l u e s

f o r a n y g i v e n a > 0 s i n c e ∣ Z ∣= ℵ_{0} .

(2) I f a = 0, S i s u n d e f i n e d s i n c e t h e

g r a p h o f y = l n x i s d i s c o n t i n u o u s a t

x = 0 . H o w e v e r, \lim_{x \to 0^{+}} l n x = - \infty

\Rightarrow \lim_{a \to 0^{+}} S = \lim_{a \to 0^{+}} \frac{i}{l n a} (n π + 0.5 π {(- 1)}^{n}) = \frac{i}{- \infty} (n π + 0.5 {(- 1)}^{n} π) = 0 \times i = 0 .

(3) I f a < 0 \Rightarrow l n a = l n (- r) = l n ({r e}^{(2 w + 1) π i})

l n a = l n r + (2 w + 1) π i w h e r e r > 0 a n d w \in Z .

∴ S = \frac{i (n π + 0.5 π {(- 1)}^{n})}{(l n r + (2 w + 1) π i)} = \frac{i (n π + 0.5 π {(- 1)}^{n})}{(l n (- a) + (2 w + 1) π i)}

S = \frac{i (l n (- a) - (2 w + 1) π i) (n π + 0.5 π {(- 1)}^{n})}{({l n (- a)}^{2} + {(2 w + 1)}^{2} π^{2})}

S = \frac{((2 w + 1) π + i l n (- a)) (n π + 0.5 π {(- 1)}^{n})}{2 ({l n}^{2} (- a) + {(2 w + 1)}^{2} π^{2})} (n, w \in Z)

I n t h i s c a s e, S i s n o t u n i q u e f o r a g i v e n a < 0,

s i m i l a r t o w h e n a > 0 .

F o r a g i v e n a < 0, S t a k e s i n f i n i t e l y

m a n y v a l u e s d e t e r m i n e d b y t h e i n t e g e r

p a i r s (n, w) w h e r e n = w o r n \neq w .

(4) F o r a \in C \Rightarrow l e t a = {c e}^{i θ} w h e r e c > 0

a n d θ = a r g (a) = a r g u m e n t o f a .

∴ l n a = {l n c e}^{i θ} = l n c + i θ

∴ S = \frac{i (n π + 0.5 {(- 1)}^{n} π)}{l n c + i θ}

S = \frac{π i (l n c - i θ) (n + 0.5 {(- 1)}^{n})}{{l n c}^{2} + θ^{2}}

S = \frac{π (n + 0.5 {(- 1)}^{n}) (θ + i l n c)}{{l n c}^{2} + θ^{2}} (n \in Z)

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