Question Number 1895 by Yozzy last updated on 22/Oct/15
$${Let}\:{us}\:{generalise}\:{the}\:{result}\:{of}\:{taking}\:{the}\:{inverse}\:{tangent}\:{of}\:{a}\:{complex}\:{number} \\ $$$${to}\:{the}\:{form}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib} \\ $$$${where}\:{a},{b},{c},{d}\in\mathbb{R}\:{and}\:{i}=\sqrt{−\mathrm{1}}.\:{Determine}\:{a}\:{and}\:{b}\:{respectively}\:{in}\:{terms} \\ $$$${of}\:{c}\:{and}\:{d}.\: \\ $$
Commented by Rasheed Soomro last updated on 23/Oct/15
$${tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib} \\ $$$$\Leftrightarrow\:{c}+{id}={tan}\left({a}+{ib}\right)=\frac{{tan}\:{a}+{tan}\:{ib}}{\mathrm{1}−{tan}\:{a}.\:{tan}\:{ib}} \\ $$$${Now}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}\:{z}\:=\frac{{e}^{{iz}} −{e}^{−{iz}} }{{e}^{{iz}} +{e}^{−{iz}} }\:,\:{z}\in\mathbb{C} \\ $$$${So},\:\:\:\:\:\:\:\:\:\:{tan}\:{ib}\:=\:\frac{{e}^{{i}\left({ib}\right)} −{e}^{−{i}\left({ib}\right)} }{{e}^{{i}\left({ib}\right)} +{e}^{−{i}\left({ib}\right)} }\:=\frac{{e}^{−{b}} −{e}^{{b}} }{{e}^{−{b}} +{e}^{{b}} } \\ $$$${Clearly}\:\:'\:{tan}\:{ib}'\:{or}'\:\:\frac{{e}^{−{b}} −{e}^{{b}} }{{e}^{−{b}} +{e}^{{b}} }\:'\:{is}\:{real}\:,\:{because}\:{b}\in\mathbb{R}. \\ $$$${And}\:\:{that}\:\:{means}\:\:{tan}\left({a}+{ib}\right)\:{is}\:{real}. \\ $$$${So}\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{c}+{id}={tan}\left({a}+{ib}\right)\Rightarrow{c}={tan}\left({a}+{ib}\right)\:\:\wedge\:{d}=\mathrm{0}\:!!! \\ $$$${But}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{c}+{id}\:{is}\:{given}\:{and}\:{d}\:{is}\:{not}\:{necessarly}\:{zero}. \\ $$$${I}−{E}\:\:\:{if}\:{a}+{ib}\:\left({tan}\left({c}+{id}\right)\:{is}\:{given}\:{then}\:{d}=\mathrm{0}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{and}\:{if}\:{c}+{id}\:{is}\:{given}\:{with}\:{d}\neq\mathrm{0}….. \\ $$$${If}\:{you}\:{see}\:{any}\:{logical}\:{defect}\:{pl}\:{inform}\:{me}. \\ $$$${Continue} \\ $$
Commented by Yozzy last updated on 23/Oct/15
$${I}\:{appreciate}\:{that}\:{you}'{ve}\:{attempted}\:{the}\:{problem}\:.\:{Now}\:{I}\:{can}\:{therefore}\:{discuss}\:{it}. \\ $$$${I}\:{believe}\:{though}\:{that}\:{tanib}\:{is}\:{not}\:{necessarily}\:{real}\:{because}\:{of}\:{the}\:{following}\:{bit} \\ $$$${of}\:{mathematics}.\: \\ $$$${By}\:{definition}\:{tanx}=\frac{{sinx}}{{cosx}}.\:{So},\:{tanib}=\frac{{sinib}}{{cosib}}. \\ $$$${The}\:{series}\:{expansions}\:{for}\:{sinx},{sinhx},{coshx}\:{and}\:{cosx}\:{help}\:{here}.\: \\ $$$${sinx}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+…\:\forall{x} \\ $$$$\therefore\:{sinix}={ix}−\frac{\left({ix}\right)^{\mathrm{3}} }{\mathrm{3}!}+\frac{\left({ix}\right)^{\mathrm{5}} }{\mathrm{5}!}−\frac{\left({ix}\right)^{\mathrm{7}} }{\mathrm{7}!}+… \\ $$$${sinix}={ix}+\frac{{ix}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{ix}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{ix}^{\mathrm{7}} }{\mathrm{7}!}+…={i}\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+…\right) \\ $$$${Since}\:{sinhx}={x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+…\:\forall{x} \\ $$$$\Rightarrow{sinix}={isinhx} \\ $$$${cosx}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+…\:\forall{x} \\ $$$$\therefore\:{cosix}=\mathrm{1}−\frac{{i}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{i}^{\mathrm{4}} {x}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{i}^{\mathrm{6}} {x}^{\mathrm{6}} }{\mathrm{6}!}+… \\ $$$${cosix}=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+… \\ $$$${But}\:{coshx}=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+…\:\forall{x} \\ $$$$\therefore\:{cosix}={coshx} \\ $$$${Thus},\:{tanib}=\frac{{isinhb}}{{coshb}}=\frac{{i}\left({e}^{{b}} −{e}^{−{b}} \right)}{{e}^{{b}} +{e}^{−{b}} }. \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 23/Oct/15
$${Thanks}\:{very}\:{much}! \\ $$$${You}\:{are}\:{very}\:{right}!\:{Actually}\:{the}\:{formula}\: \\ $$$${of}\:\:{tan}\:{z}\:\left(\frac{{e}^{{iz}} −{e}^{−{iz}} \:}{{e}^{{iz}} +{e}^{−{iz}} }\right)\:\:{was}\:{misprinted}\:{in}\:{a}\:{book}! \\ $$$${The}\:{book}\:{was}\:{not}\:{low}\:{standard}\:{but}\:{printing} \\ $$$${errors}\:{are}\:{common}\:{here}.\:{I}\:{used}\:{this}\:{wrong} \\ $$$${formula}\:{and}\:{reached}\:{at}\:{wrong}\:{result}! \\ $$$${Anyway}\:{thanks}\:{again}\:{to}\:{correct}\:{my}\:{mistake}.\: \\ $$
Commented by Yozzy last updated on 23/Oct/15
$${No}\:{problem}.\:{I}'{m}\:{here}\:{to}\:{learn}\:{and}\:{help}\:{too}. \\ $$
Commented by Rasheed Soomro last updated on 23/Oct/15
$${Thank}\boldsymbol{{S}}. \\ $$
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) −−−−−−−−−−−−−−−−−−−](https://www.tinkutara.com/question/Q1905.png)
$${tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib}\: \\ $$$${Any}\:{of}\:\:{a}\:\:{and}\:\:{b}\:\:{depends}\:{on}\:{both}\:{c}\:{and}\:{d}: \\ $$$$\:\:\:\:\:\:\:{a}\:=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{c}}{\mathrm{1}−\mathrm{c}^{\mathrm{2}} −\mathrm{d}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{2}{i}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{id}}{\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right) \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$\boldsymbol{\mathrm{Proof}}: \\ $$$${tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib}\:…………………………\left(\mathrm{1}\right) \\ $$$${tan}^{−\mathrm{1}} \left({c}−{id}\right)={a}−{ib}\:\:\:\:\left[\because\:\overline {{tan}\:{z}}={tan}\:\overline {{z}}\:\:\right]……\left(\mathrm{2}\right) \\ $$$${Adding}\:\left(\mathrm{1}\right)\:{and}\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{a}={tan}^{−\mathrm{1}} \left({c}+{id}\right)+{tan}^{−\mathrm{1}} \left({c}−{id}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={tan}^{−\mathrm{1}} \frac{\left({c}+{id}\right)+\left({c}−{id}\right)}{\mathrm{1}−\left({c}+{id}\right)\left({c}−{id}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:={tan}^{−\mathrm{1}} \frac{\mathrm{2}{c}}{\mathrm{1}−\left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)} \\ $$$${Or}\:{a}\:=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{c}}{\mathrm{1}−\mathrm{c}^{\mathrm{2}} −\mathrm{d}^{\mathrm{2}} }\right)……………..{I}\left({Proved}\right) \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${Similarly}\:{subtracting}\:\left(\mathrm{2}\right)\:{from}\:\left(\mathrm{1}\right),\:{we}\:{get} \\ $$$$\:\:\mathrm{2}{ib}={tan}^{−\mathrm{1}} \left({c}+{id}\right)−{tan}^{−\mathrm{1}} \left({c}−{id}\right) \\ $$$$\:\:\:\:\:\:\:\:\:={tan}^{−\mathrm{1}} \frac{\left({c}+{id}\right)−\left({c}−{id}\right)}{\mathrm{1}+\left({c}+{id}\right)\left({c}−{id}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:={tan}^{−\mathrm{1}} \frac{\mathrm{2}{id}}{\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$$\:\:\:\:{tan}\:\mathrm{2}{ib}=\frac{\mathrm{2}{id}}{\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\mathrm{2}{ib}={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{id}}{\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{2}{i}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{id}}{\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }\right)……………….{II}\left(\:{Proved}\right) \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$$ \\ $$
Commented by Yozzy last updated on 23/Oct/15
$${I}\:{would}\:{have}\:{appreciated}\:{a}\:{proof}\:{of}\:{your}\:{answer}. \\ $$
Commented by Yozzy last updated on 23/Oct/15
$${Some}\:{calculations}\:{I}'{ve}\:{done}\:{has}\:{interestingly}\:{showed}\:{that}\:{for}\:{c}=\mathrm{1}\:{and}\:{d}=\mathrm{1} \\ $$$${my}\:{result}\:{yields}\:{both}\:{your}\:{value}\:{for}\:{a}\:{and}\:{Wolfram}\:{Alpha}'{s}\:{answer}\:{for}\:{a}. \\ $$$${Two}\:{different}\:{answers}\:{for}\:{a}\:{were}\:{obtained}!\:{How}\:{does}\:{one}\:{decide}\:{which} \\ $$$${value}\:{to}\:{take}? \\ $$
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))})))](https://www.tinkutara.com/question/Q1907.png)
$${tan}^{−\mathrm{1}} \left({c}+{id}\right)={a}+{ib}\:\:{a},{b},{c},{d}\in\mathbb{R}. \\ $$$$\therefore\:{c}+{id}={tan}\left({a}+{ib}\right)=\frac{{tana}+{tanib}}{\mathrm{1}−{tanatanib}} \\ $$$${Now},\:{tanib}={itanhb}. \\ $$$$\therefore\:{c}+{id}=\frac{{tana}+{itanhb}}{\mathrm{1}−{itanatanhb}}\Rightarrow\left({c}+{id}\right)\left(\mathrm{1}−{itanatanhb}\right)={tana}+{itanhb} \\ $$$${c}+{dtanatanhb}+{i}\left({d}−{ctanatanhb}\right)={tana}+{itanhb} \\ $$$${Equating}\:{real}\:{and}\:{imaginary}\:{parts}\:{we}\:{get}\: \\ $$$$\:\:{c}+{dtanatanhb}={tana}\Rightarrow{tana}\left(\mathrm{1}−{dtanhb}\right)={c}\Rightarrow{tana}=\frac{{c}}{\mathrm{1}−{dtanhb}}…….\left({i}\right) \\ $$$${And} \\ $$$$\:\:{tanhb}={d}−{ctanatanhb}\Rightarrow{tana}=\frac{{d}−{tanhb}}{{ctanhb}}……….\left({ii}\right) \\ $$$${Since}\:\left({i}\right)=\left({ii}\right)\Rightarrow\:\frac{{c}}{\mathrm{1}−{dtanhb}}=\frac{{d}−{tanhb}}{{ctanhb}}.\:{Let}\:\omega={tanhb}. \\ $$$$\therefore\:\frac{{c}}{\mathrm{1}−{d}\omega}=\frac{{d}−\omega}{{c}\omega}\Rightarrow{c}^{\mathrm{2}} \omega=\left(\mathrm{1}−{d}\omega\right)\left({d}−\omega\right)={d}−\omega−{d}^{\mathrm{2}} \omega+{d}\omega^{\mathrm{2}} \\ $$$${d}\omega^{\mathrm{2}} −\left({d}^{\mathrm{2}} +\mathrm{1}+{c}^{\mathrm{2}} \right)\omega+{d}=\mathrm{0} \\ $$$${This}\:{is}\:{a}\:{quadratic}\:{so} \\ $$$$\:\:\:\:\omega=\frac{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}{d}}\:\:\:\: \\ $$$$\therefore\:\:{tanhb}=\frac{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}{d}}\:\:\:\left({d}\neq\mathrm{0}\right)\:{and}\:{certainly}\:\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} >\mathrm{4}{d}\:\:\forall{c},{d}\in\mathbb{R}\: \\ $$$$ \\ $$$${Let}\:{y}={tanh}^{−\mathrm{1}} {x}.\Rightarrow{tanhy}={x}\Rightarrow{sinhy}={xcoshy} \\ $$$${e}^{{y}} −{e}^{−{y}} ={xe}^{{y}} +{xe}^{−{y}} \\ $$$$×{e}^{{y}} :\:{e}^{\mathrm{2}{y}} −\mathrm{1}={xe}^{\mathrm{2}{y}} +{x} \\ $$$${e}^{\mathrm{2}{y}} \left(\mathrm{1}−{x}\right)={x}+\mathrm{1}\Rightarrow{e}^{\mathrm{2}{y}} =\frac{{x}+\mathrm{1}}{\mathrm{1}−{x}}\Rightarrow\mathrm{2}{y}={ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)\:\mid{x}\mid<\mathrm{1}\:\left({so}\:{y}\:{is}\:{real}\:{and}\:{defined}\right) \\ $$$${So}\:{b}={tanh}^{−\mathrm{1}} \left[\frac{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}{d}}\right]\:{with}\:{b}\in\mathbb{R}\:{only}\:{if}\:\mid\frac{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}{d}}\mid<\mathrm{1} \\ $$$$\Rightarrow\mid\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}\mid<\mathrm{2}\mid{d}\mid.\:{If}\:{we}\:{assume}\:{this}\:{condition}\:{is}\:{satisfied} \\ $$$${we}\:{can}\:{find}\:{a}. \\ $$$$ \\ $$$${dtanhb}=\frac{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}} \\ $$$$\mathrm{1}−{dtanhb}=\frac{\mathrm{2}−\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}}{\mathrm{2}} \\ $$$$\therefore\:{tana}=\frac{\mathrm{2}{c}}{\mathrm{2}−\left\{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}\right\}} \\ $$$${tana}\in\left(−\infty,+\infty\right)\:{so}\:{c}\:{and}\:{d}\:{can}\:{be}\:{such}\:{that}\:{the}\:{above}\:{equation}\:{is}\:{valid} \\ $$$${for}\:{any}\:{real}\:{c}\:{and}\:{d}\:{satisfying}\:{the}\:{condition}\:\mid\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}\mid<\mathrm{2}\mid{d}\mid \\ $$$$\therefore\:{a}={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{c}}{\mathrm{2}−\left\{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{1}+{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{d}}\right\}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 23/Oct/15
$$\boldsymbol{\mathrm{Very}}\:\boldsymbol{\mathrm{intresting}}!\:\boldsymbol{\mathrm{Your}}\:\boldsymbol{\mathrm{work}}\:\boldsymbol{\mathrm{needs}}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{deep}}\:\boldsymbol{\mathrm{study}}….\: \\ $$
Commented by Rasheed Soomro last updated on 23/Oct/15
$${That}\:{means}\:{you}\:{are}\:{only}\:{undergraduate}!\:{I}\:{appreciate}\: \\ $$$${your}\:{deep}\:{approach}\:{at}\:{this}\:{early}\:{stage}\:{and}\:{offer}\:{good} \\ $$$${wishes}\:{for}\:{your}\:{future}! \\ $$
Commented by Yozzy last updated on 23/Oct/15
$${Yes}!\:{I}\:{know}\:{this}\:{isn}'{t}\:{perfect}\:{since}\:{this}\:{doesn}'{t}\:{give}\:{a}\:{complete}\:{picture}\:{I}\:{think} \\ $$$${of}\:{the}\:{inverse}\:{tangent}\:{of}\:{complex}\:{numbers}.\: \\ $$$${I}\:{also}\:{haven}'{t}\:{started}\:{university}\:{as}\:{yet}\:{so}\:{I}\:{don}'{t}\:{know}\:{much}\:{of}\:{complex} \\ $$$${analysis}.\:{I}\:{begin}\:{next}\:{year}. \\ $$$$ \\ $$
Commented by 123456 last updated on 23/Oct/15
$$\left.\mathrm{good}\:\mathrm{lucky}\::\right) \\ $$
Commented by Yozzy last updated on 24/Oct/15
$${Thank}\:{you}!\::\mathrm{D}\: \\ $$