I-need-the-sol-plz-find-the-imaginary-and-real-parts-of-log-sin-a-ib-

Question Number 73913 by arkanmath7@gmail.com last updated on 16/Nov/19

I n e e d t h e s o l . p l z

f i n d t h e i m a g i n a r y a n d r e a l p a r t s o f

l o g s i n (a + i b) ?

Answered by Tanmay chaudhury last updated on 16/Nov/19

L o g (s i n a c o s i b + c o s a s i n i b)

L o g (s i n a . c o s h b + c o s a . i s i n h b)

R c o s α = s i n a . c o s h b

R s i n α = c o s a . s i n h b

L o g (R c o s α + i R s i n α)

L o g [{R e}^{i α}] ⇛ L o g [{R e}^{i (2 n π + α)}]

L o g [{R e}^{i (2 n π + α)}] = l o g R + i (2 n π + α)

R = {[{(s i n a . c o s h b)}^{2} + {(c o s a . s i n h b)}^{2}]}^{\frac{1}{2}}

t a n α = \frac{c o s a . s i n h b}{s i n a . c o s h b} ⇛ α = {t a n}^{- 1} (\frac{c o s a . s i n h b}{s i n a . c o s h b})

Commented by arkanmath7@gmail.com last updated on 16/Nov/19

t h n x

Answered by Smail last updated on 16/Nov/19

l n (s i n (a + i b)) = x + i y

s i n (a + i b) = e^{x} e^{i y}

s i n (a) c o s (i b) + s i n (i b) c o s (a) = e^{x} e^{i y}

s i n (a) c o s h (b) + i s i n h (b) c o s (a) = e^{x} e^{i y}

\frac{s i n h (b) c o s (a)}{s i n (a) c o s h (b)} = t a n (y)

y = {t a n}^{- 1} (t a n h (b) c o t (a)))

e^{x} = \sqrt{{s i n}^{2} (a) {c o s h}^{2} (b) + {s i n h}^{2} (b) {c o s}^{2} (a)}

x = \frac{1}{2} l n ({s i n}^{2} (a) {c o s h}^{2} (b) + {s i n h}^{2} (b) {c o s}^{2} (a))

s i n (a + i b) = \frac{1}{2} l n ({s i n}^{2} (a) {c o s h}^{2} (b) + {s i n h}^{2} (b) {c o s}^{2} (a)) + {i t a n}^{- 1} (t a n h (b) c o t (a))

Commented by Smail last updated on 16/Nov/19

l n (s i n (a + i b)) = \frac{1}{2} l n ({s i n}^{2} (a) {c o s h}^{2} (b) + {s i n h}^{2} (b) {c o s}^{2} (a)) + {i t a n}^{- 1} (t a n h (b) c o t (a))

Facebook Tweet Pin