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-an

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

tan^(−1) (c+id)=a+ib ⇔ c+id=tan(a+ib)=((tan a+tan ib)/(1−tan a. tan ib)) Now, tan z =((e^(iz) −e^(−iz) )/(e^(iz) +e^(−iz) )) , z∈C So, tan ib = ((e^(i(ib)) −e^(−i(ib)) )/(e^(i(ib)) +e^(−i(ib)) )) =((e^(−b) −e^b )/(e^(−b) +e^b )) Clearly ′ tan ib′ or′ ((e^(−b) −e^b )/(e^(−b) +e^b )) ′ is real , because b∈R. And that means tan(a+ib) is real. So , c+id=tan(a+ib)⇒c=tan(a+ib) ∧ d=0 !!! But c+id is given and d is not necessarly zero. I−E if a+ib (tan(c+id) is given then d=0 and if c+id is given with d≠0..... If you see any logical defect pl inform me. Continue

{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 \dots . .

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 appreciate that you′ve attempted the problem . Now I can therefore discuss it. I believe though that tanib is not necessarily real because of the following bit of mathematics. By definition tanx=((sinx)/(cosx)). So, tanib=((sinib)/(cosib)). The series expansions for sinx,sinhx,coshx and cosx help here. sinx=x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+... ∀x ∴ sinix=ix−(((ix)^3 )/(3!))+(((ix)^5 )/(5!))−(((ix)^7 )/(7!))+... sinix=ix+((ix^3 )/(3!))+((ix^5 )/(5!))+((ix^7 )/(7!))+...=i(x+(x^3 /(3!))+(x^5 /(5!))+(x^7 /(7!))+...) Since sinhx=x+(x^3 /(3!))+(x^5 /(5!))+(x^7 /(7!))+... ∀x ⇒sinix=isinhx cosx=1−(x^2 /(2!))+(x^4 /(4!))−(x^6 /(6!))+... ∀x ∴ cosix=1−((i^2 x^2 )/(2!))+((i^4 x^4 )/(4!))−((i^6 x^6 )/(6!))+... cosix=1+(x^2 /(2!))+(x^4 /(4!))+(x^6 /(6!))+... But coshx=1+(x^2 /(2!))+(x^4 /(4!))+(x^6 /(6!))+... ∀x ∴ cosix=coshx Thus, tanib=((isinhb)/(coshb))=((i(e^b −e^(−b) ))/(e^b +e^(−b) )).

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!} + \dots \forall x

∴ s i n i x = i x - \frac{{(i x)}^{3}}{3!} + \frac{{(i x)}^{5}}{5!} - \frac{{(i x)}^{7}}{7!} + \dots

s i n i x = i x + \frac{{i x}^{3}}{3!} + \frac{{i x}^{5}}{5!} + \frac{{i x}^{7}}{7!} + \dots = i (x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + \dots)

S i n c e s i n h x = x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \frac{x^{7}}{7!} + \dots \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!} + \dots \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!} + \dots

c o s i x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \dots

B u t c o s h x = 1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \frac{x^{6}}{6!} + \dots \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

Thanks very much! You are very right! Actually the formula of tan z (((e^(iz) −e^(−iz) )/(e^(iz) +e^(−iz) ))) was misprinted in a book! The book was not low standard but printing errors are common here. I used this wrong formula and reached at wrong result! Anyway thanks again to correct my mistake.

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

No problem. I′m here to learn and help too.

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

tan^(−1) (c+id)=a+ib Any of a and b depends on both c and d: a =(1/2)tan^(−1) (((2c)/(1−c^2 −d^2 ))) b=(1/(2i))tan^(−1) (((2id)/(1+c^2 +d^2 ))) −−−−−−−−−−−−−−−−− Proof: tan^(−1) (c+id)=a+ib ..............................(1) tan^(−1) (c−id)=a−ib [∵ tan z^(−) =tan z^(−) ]......(2) Adding (1) and (2) 2a=tan^(−1) (c+id)+tan^(−1) (c−id) =tan^(−1) (((c+id)+(c−id))/(1−(c+id)(c−id))) =tan^(−1) ((2c)/(1−(c^2 +d^2 ))) Or a =(1/2)tan^(−1) (((2c)/(1−c^2 −d^2 ))).................I(Proved) −−−−−−−−−−−−−−−− Similarly subtracting (2) from (1), we get 2ib=tan^(−1) (c+id)−tan^(−1) (c−id) =tan^(−1) (((c+id)−(c−id))/(1+(c+id)(c−id))) =tan^(−1) ((2id)/(1+c^2 +d^2 )) tan 2ib=((2id)/(1+c^2 +d^2 )) 2ib=tan^(−1) (((2id)/(1+c^2 +d^2 ))) b=(1/(2i))tan^(−1) (((2id)/(1+c^2 +d^2 )))...................II( Proved) −−−−−−−−−−−−−−−−−−−

{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}})

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Proof :

{t a n}^{- 1} (c + i d) = a + i b \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots (1)

{t a n}^{- 1} (c - i d) = a - i b [∵ \overset{―}{t a n z} = t a n \overset{―}{z}] \dots \dots (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}}) \dots \dots \dots \dots \dots . . I (P r o v e d)

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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}}) \dots \dots \dots \dots \dots \dots . I I (P r o v e d)

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Commented by Yozzy last updated on 23/Oct/15

I would have appreciated a proof of your answer.

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

Some calculations I′ve done has interestingly showed that for c=1 and d=1 my result yields both your value for a and Wolfram Alpha′s answer for a. Two different answers for a were obtained! How does one decide which value to take?

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} \dots \dots . (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} \dots \dots \dots . (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....

Very intresting! Your work needs a deep study \dots .

Commented by Rasheed Soomro last updated on 23/Oct/15

That means you are only undergraduate! I appreciate your deep approach at this early stage and offer good wishes for your future!

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

Yes! I know this isn′t perfect since this doesn′t give a complete picture I think of the inverse tangent of complex numbers. I also haven′t started university as yet so I don′t know much of complex analysis. I begin next year.

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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