Let us generalise the result of taking the inverse tangent of a complex number to the form tan^(−1) (c+id)=a+ib where a,b,c,d∈R and i=(√(−1)). Determine a and b respectively in terms of c and d.


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Question Number 1895 by Yozzy last updated on 22/Oct/15

$L e t u s g e n e r a l i s e t h e r e s u l t o f t a k i n g t h e i n v e r s e t a n g e n t o f a c o m p l e x n u m b e r$ $t o t h e f o r m$ ${t a n}^{- 1} (c + i d) = a + i b$ $w h e r e a, b, c, d \in R a n d i = \sqrt{- 1} . D e t e r m i n e a a n d b r e s p e c t i v e l y i n t e r m s$ $o f c a n d d .$
Commented by Rasheed Soomro last updated on 23/Oct/15

${t a n}^{- 1} (c + i d) = a + i b$ $\Leftrightarrow c + i d = t a n (a + i b) = \frac{t a n a + t a n i b}{1 - t a n a . t a n i b}$ $N o w,$ $t a n z = \frac{e^{i z} - e^{- i z}}{e^{i z} + e^{- i z}}, z \in C$ $S o, t a n i b = \frac{e^{i (i b)} - e^{- i (i b)}}{e^{i (i b)} + e^{- i (i b)}} = \frac{e^{- b} - e^{b}}{e^{- b} + e^{b}}$ $C l e a r l y^{'} t a n {i b}^{'} {o r}^{'} \frac{e^{- b} - e^{b}}{e^{- b} + e^{b}}^{'} i s r e a l, b e c a u s e b \in R .$ $A n d t h a t m e a n s t a n (a + i b) i s r e a l .$ $S o,$ $c + i d = t a n (a + i b) \Rightarrow c = t a n (a + i b) \land d = 0!!!$ $B u t$ $c + i d i s g i v e n a n d d i s n o t n e c e s s a r l y z e r o .$ $I - E i f a + i b (t a n (c + i d) i s g i v e n t h e n d = 0$ $a n d i f c + i d i s g i v e n w i t h d \neq 0 . . . . .$ $I f y o u s e e a n y l o g i c a l d e f e c t p l i n f o r m m e .$ $C o n t i n u e$
Commented by Yozzy last updated on 23/Oct/15

$I a p p r e c i a t e t h a t {y o u}^{'} v e a t t e m p t e d t h e p r o b l e m . N o w I c a n t h e r e f o r e d i s c u s s i t .$ $I b e l i e v e t h o u g h t h a t t a n i b i s n o t n e c e s s a r i l y r e a l b e c a u s e o f t h e f o l l o w i n g b i t$ $o f m a t h e m a t i c s .$ $B y d e f i n i t i o n t a n x = \frac{s i n x}{c o s x} . S o, t a n i b = \frac{s i n i b}{c o s i b} .$ $T h e s e r i e s e x p a n s i o n s f o r s i n x, s i n h x, c o s h x a n d c o s x h e l p h e r e .$ $s i n x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + . . . \forall x$ $∴ s i n i x = i x - \frac{{(i x)}^{3}}{3!} + \frac{{(i x)}^{5}}{5!} - \frac{{(i x)}^{7}}{7!} + . . .$ $s i n i x = i x + \frac{{i x}^{3}}{3!} + \frac{{i x}^{5}}{5!} + \frac{{i x}^{7}}{7!} + . . . = i (x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + . . .)$ $S i n c e s i n h x = x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + . . . \forall x$ $\Rightarrow s i n i x = i s i n h x$ $c o s x = 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + . . . \forall x$ $∴ c o s i x = 1 - \frac{i^{2} x^{2}}{2!} + \frac{i^{4} x^{4}}{4!} - \frac{i^{6} x^{6}}{6!} + . . .$ $c o s i x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + . . .$ $B u t c o s h x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + . . . \forall x$ $∴ c o s i x = c o s h x$ $T h u s, t a n i b = \frac{i s i n h b}{c o s h b} = \frac{i (e^{b} - e^{- b})}{e^{b} + e^{- b}} .$
Commented by Rasheed Soomro last updated on 23/Oct/15

$T h a n k s v e r y m u c h!$ $Y o u a r e v e r y r i g h t! A c t u a l l y t h e f o r m u l a$ $o f t a n z (\frac{e^{i z} - e^{- i z}}{e^{i z} + e^{- i z}}) w a s m i s p r i n t e d i n a b o o k!$ $T h e b o o k w a s n o t l o w s t a n d a r d b u t p r i n t i n g$ $e r r o r s a r e c o m m o n h e r e . I u s e d t h i s w r o n g$ $f o r m u l a a n d r e a c h e d a t w r o n g r e s u l t!$ $A n y w a y t h a n k s a g a i n t o c o r r e c t m y m i s t a k e .$
Commented by Yozzy last updated on 23/Oct/15

$N o p r o b l e m . I^{'} m h e r e t o l e a r n a n d h e l p t o o .$
Commented by Rasheed Soomro last updated on 23/Oct/15

$T h a n k S .$
	Answered by Rasheed Soomro last updated on 25/Oct/15

	${t a n}^{- 1} (c + i d) = a + i b$ $A n y o f a a n d b d e p e n d s o n b o t h c a n d d :$ $a = \frac{1}{2} {t a n}^{- 1} (\frac{2 c}{1 - c^{2} - d^{2}})$ $b = \frac{1}{2 i} {t a n}^{- 1} (\frac{2 i d}{1 + c^{2} + d^{2}})$ $- - - - - - - - - - - - - - - - -$ $Proof :$ ${t a n}^{- 1} (c + i d) = a + i b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)$ ${t a n}^{- 1} (c - i d) = a - i b [∵ \overset{―}{t a n z} = t a n \overset{―}{z}] . . . . . . (2)$ $A d d i n g (1) a n d (2)$ $2 a = {t a n}^{- 1} (c + i d) + {t a n}^{- 1} (c - i d)$ $= {t a n}^{- 1} \frac{(c + i d) + (c - i d)}{1 - (c + i d) (c - i d)}$ $= {t a n}^{- 1} \frac{2 c}{1 - (c^{2} + d^{2})}$ $O r a = \frac{1}{2} {t a n}^{- 1} (\frac{2 c}{1 - c^{2} - d^{2}}) . . . . . . . . . . . . . . . . . I (P r o v e d)$ $- - - - - - - - - - - - - - - -$ $S i m i l a r l y s u b t r a c t i n g (2) f r o m (1), w e g e t$ $2 i b = {t a n}^{- 1} (c + i d) - {t a n}^{- 1} (c - i d)$ $= {t a n}^{- 1} \frac{(c + i d) - (c - i d)}{1 + (c + i d) (c - i d)}$ $= {t a n}^{- 1} \frac{2 i d}{1 + c^{2} + d^{2}}$ $t a n 2 i b = \frac{2 i d}{1 + c^{2} + d^{2}}$ $2 i b = {t a n}^{- 1} (\frac{2 i d}{1 + c^{2} + d^{2}})$ $b = \frac{1}{2 i} {t a n}^{- 1} (\frac{2 i d}{1 + c^{2} + d^{2}}) . . . . . . . . . . . . . . . . . . . I I (P r o v e d)$ $- - - - - - - - - - - - - - - - - - -$
	Commented by Yozzy last updated on 23/Oct/15

	$I w o u l d h a v e a p p r e c i a t e d a p r o o f o f y o u r a n s w e r .$
	Commented by Yozzy last updated on 23/Oct/15

	$S o m e c a l c u l a t i o n s I^{'} v e d o n e h a s i n t e r e s t i n g l y s h o w e d t h a t f o r c = 1 a n d d = 1$ $m y r e s u l t y i e l d s b o t h y o u r v a l u e f o r a a n d W o l f r a m {A l p h a}^{'} s a n s w e r f o r a .$ $T w o d i f f e r e n t a n s w e r s f o r a w e r e o b t a i n e d! H o w d o e s o n e d e c i d e w h i c h$ $v a l u e t o t a k e ?$
	Answered by Yozzy last updated on 23/Oct/15
	$tan^(−1) (c+id)=a+ib a,b,c,d∈R. ∴ c+id=tan(a+ib)=((tana+tanib)/(1−tanatanib)) Now, tanib=itanhb. ∴ c+id=((tana+itanhb)/(1−itanatanhb))⇒(c+id)(1−itanatanhb)=tana+itanhb c+dtanatanhb+i(d−ctanatanhb)=tana+itanhb Equating real and imaginary parts we get c+dtanatanhb=tana⇒tana(1−dtanhb)=c⇒tana=(c/(1−dtanhb)).......(i) And tanhb=d−ctanatanhb⇒tana=((d−tanhb)/(ctanhb))..........(ii) Since (i)=(ii)⇒ (c/(1−dtanhb))=((d−tanhb)/(ctanhb)). Let ω=tanhb. ∴ (c/(1−dω))=((d−ω)/(cω))⇒c^2 ω=(1−dω)(d−ω)=d−ω−d^2 ω+dω^2 dω^2 −(d^2 +1+c^2 )ω+d=0 This is a quadratic so ω=(((1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/(2d)) ∴ tanhb=(((1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/(2d)) (d≠0) and certainly (1+c^2 +d^2 )^2 >4d ∀c,d∈R Let y=tanh^(−1) x.⇒tanhy=x⇒sinhy=xcoshy e^y −e^(−y) =xe^y +xe^(−y) ×e^y : e^(2y) −1=xe^(2y) +x e^(2y) (1−x)=x+1⇒e^(2y) =((x+1)/(1−x))⇒2y=ln(((1+x)/(1−x)))⇒y=(1/2)ln(((1+x)/(1−x))) ∣x∣<1 (so y is real and defined) So b=tanh^(−1) [(((1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/(2d))] with b∈R only if ∣(((1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/(2d))∣<1 ⇒∣(1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d))∣<2∣d∣. If we assume this condition is satisfied we can find a. dtanhb=(((1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/2) 1−dtanhb=((2−(1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d)))/2) ∴ tana=((2c)/(2−{(1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d))})) tana∈(−∞,+∞) so c and d can be such that the above equation is valid for any real c and d satisfying the condition ∣(1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d))∣<2∣d∣ ∴ a=tan^(−1) (((2c)/(2−{(1+c^2 +d^2 )±(√((1+c^2 +d^2 )^2 −4d))})))$
	${t a n}^{- 1} (c + i d) = a + i b a, b, c, d \in R .$ $∴ c + i d = t a n (a + i b) = \frac{t a n a + t a n i b}{1 - t a n a t a n i b}$ $N o w, t a n i b = i t a n h b .$ $∴ c + i d = \frac{t a n a + i t a n h b}{1 - i t a n a t a n h b} \Rightarrow (c + i d) (1 - i t a n a t a n h b) = t a n a + i t a n h b$ $c + d t a n a t a n h b + i (d - c t a n a t a n h b) = t a n a + i t a n h b$ $E q u a t i n g r e a l a n d i m a g i n a r y p a r t s w e g e t$ $c + d t a n a t a n h b = t a n a \Rightarrow t a n a (1 - d t a n h b) = c \Rightarrow t a n a = \frac{c}{1 - d t a n h b} . . . . . . . (i)$ $A n d$ $t a n h b = d - c t a n a t a n h b \Rightarrow t a n a = \frac{d - t a n h b}{c t a n h b} . . . . . . . . . . (i i)$ $S i n c e (i) = (i i) \Rightarrow \frac{c}{1 - d t a n h b} = \frac{d - t a n h b}{c t a n h b} . L e t ω = t a n h b .$ $∴ \frac{c}{1 - d ω} = \frac{d - ω}{c ω} \Rightarrow c^{2} ω = (1 - d ω) (d - ω) = d - ω - d^{2} ω + d ω^{2}$ $d ω^{2} - (d^{2} + 1 + c^{2}) ω + d = 0$ $T h i s i s a q u a d r a t i c s o$ $ω = \frac{(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2 d}$ $∴ t a n h b = \frac{(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2 d} (d \neq 0) a n d c e r t a i n l y {(1 + c^{2} + d^{2})}^{2} > 4 d \forall c, d \in R$ $L e t y = {t a n h}^{- 1} x . \Rightarrow t a n h y = x \Rightarrow s i n h y = x c o s h y$ $e^{y} - e^{- y} = {x e}^{y} + {x e}^{- y}$ $\times e^{y} : e^{2 y} - 1 = {x e}^{2 y} + x$ $e^{2 y} (1 - x) = x + 1 \Rightarrow e^{2 y} = \frac{x + 1}{1 - x} \Rightarrow 2 y = l n (\frac{1 + x}{1 - x}) \Rightarrow y = \frac{1}{2} l n (\frac{1 + x}{1 - x}) ∣ x ∣< 1 (s o y i s r e a l a n d d e f i n e d)$ $S o b = {t a n h}^{- 1} [\frac{(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2 d}] w i t h b \in R o n l y i f ∣ \frac{(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2 d} ∣< 1$ $\Rightarrow∣ (1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d} ∣< 2 ∣ d ∣ . I f w e a s s u m e t h i s c o n d i t i o n i s s a t i s f i e d$ $w e c a n f i n d a .$ $d t a n h b = \frac{(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2}$ $1 - d t a n h b = \frac{2 - (1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}{2}$ $∴ t a n a = \frac{2 c}{2 - {(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}}$ $t a n a \in (- \infty, + \infty) s o c a n d d c a n b e s u c h t h a t t h e a b o v e e q u a t i o n i s v a l i d$ $f o r a n y r e a l c a n d d s a t i s f y i n g t h e c o n d i t i o n ∣ (1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d} ∣< 2 ∣ d ∣$ $∴ a = {t a n}^{- 1} (\frac{2 c}{2 - {(1 + c^{2} + d^{2}) \pm \sqrt{{(1 + c^{2} + d^{2})}^{2} - 4 d}}})$
	Commented by Rasheed Soomro last updated on 23/Oct/15

	$Very intresting! Your work needs a deep study . . . .$
	Commented by Rasheed Soomro last updated on 23/Oct/15

	$T h a t m e a n s y o u a r e o n l y u n d e r g r a d u a t e! I a p p r e c i a t e$ $y o u r d e e p a p p r o a c h a t t h i s e a r l y s t a g e a n d o f f e r g o o d$ $w i s h e s f o r y o u r f u t u r e!$
	Commented by Yozzy last updated on 23/Oct/15

	$Y e s! I k n o w t h i s {i s n}^{'} t p e r f e c t s i n c e t h i s {d o e s n}^{'} t g i v e a c o m p l e t e p i c t u r e I t h i n k$ $o f t h e i n v e r s e t a n g e n t o f c o m p l e x n u m b e r s .$ $I a l s o {h a v e n}^{'} t s t a r t e d u n i v e r s i t y a s y e t s o I {d o n}^{'} t k n o w m u c h o f c o m p l e x$ $a n a l y s i s . I b e g i n n e x t y e a r .$
	Commented by 123456 last updated on 23/Oct/15

	$good lucky :)$
	Commented by Yozzy last updated on 24/Oct/15

	$T h a n k y o u! : D$
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