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Question Number 116560 by Bird last updated on 04/Oct/20

if arctan(x+iy) =a+ib  with a and b reals determine  a and b

$${if}\:{arctan}\left({x}+{iy}\right)\:={a}+{ib} \\ $$$${with}\:{a}\:{and}\:{b}\:{reals}\:{determine} \\ $$$${a}\:{and}\:{b} \\ $$

Commented by MJS_new last updated on 05/Oct/20

arctan (x+iy) =a+ib  a=(π/2)sign x +(1/2)(arctan ((y−1)/x) −arctan ((y+1)/x))  b=(1/4)ln ((x^2 +(y+1)^2 )/(x^2 +(y−1)^2 ))  if x=0  a=real ((i/2)ln ((1+y)/(1−y)))  b=imag ((i/2)ln ((1+y)/(1−y)))

$$\mathrm{arctan}\:\left({x}+\mathrm{i}{y}\right)\:={a}+\mathrm{i}{b} \\ $$$${a}=\frac{\pi}{\mathrm{2}}\mathrm{sign}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{arctan}\:\frac{{y}−\mathrm{1}}{{x}}\:−\mathrm{arctan}\:\frac{{y}+\mathrm{1}}{{x}}\right) \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{if}\:{x}=\mathrm{0} \\ $$$${a}=\mathrm{real}\:\left(\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}\right) \\ $$$${b}=\mathrm{imag}\:\left(\frac{\mathrm{i}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+{y}}{\mathrm{1}−{y}}\right) \\ $$

Answered by Olaf last updated on 05/Oct/20

x+iy = tan(a+ib)  x+iy = ((tana+tanib)/(1−tanatanib))    tanib = ((sinib)/(cosib))  tanib = (((e^(i(ib)) −e^(−i(ib)) )/(2i))/((e^(i(ib)) +e^(−i(ib)) )/2)) = −i((e^(−b) −e^(ib) )/(e^(−ib) +e^(ib) ))  tanib  = i((e^b −e^(−ib) )/(e^(ib) +e^(−ib) )) = i((sinhb)/(coshb)) = itanhb    x+iy = ((tana+itanhb)/(1−itanatanhb))  x+iy = (((tana+itanhb)(1+itanatanhb))/(1+tan^2 atanh^2 b))  x+iy = ((tana(1−tanh^2 b)+itanhb(1+tan^2 a))/(1+tan^2 atanh^2 b))    (y/x) = ((tanhb(1+tan^2 a))/(tana(1−tanh^2 b)))  (y/x) = ((tanhbcosh^2 b)/(tanacos^2 a)) = ((sinhbcoshb)/(sinacosa))  (y/x) = ((tanhbcosh^2 b)/(tanacos^2 a)) = ((sinh2b)/(sin2a)) (1)    x^2 +y^2  = ((tan^2 a)/(cosh^4 b))+((tanh^2 b)/(cos^4 a))  x^2 +y^2  = ((sin^2 acos^2 a+sinh^2 bcosh^2 b)/(cos^4 acosh^4 b))  x^2 +y^2  = ((sin^2 2a+sinh^2 2b)/(4cos^4 acosh^4 b)) (2)  work in progress...

$${x}+{iy}\:=\:\mathrm{tan}\left({a}+{ib}\right) \\ $$$${x}+{iy}\:=\:\frac{\mathrm{tan}{a}+\mathrm{tan}{ib}}{\mathrm{1}−\mathrm{tan}{a}\mathrm{tan}{ib}} \\ $$$$ \\ $$$$\mathrm{tan}{ib}\:=\:\frac{\mathrm{sin}{ib}}{\mathrm{cos}{ib}} \\ $$$$\mathrm{tan}{ib}\:=\:\frac{\frac{{e}^{{i}\left({ib}\right)} −{e}^{−{i}\left({ib}\right)} }{\mathrm{2}{i}}}{\frac{{e}^{{i}\left({ib}\right)} +{e}^{−{i}\left({ib}\right)} }{\mathrm{2}}}\:=\:−{i}\frac{{e}^{−{b}} −{e}^{{ib}} }{{e}^{−{ib}} +{e}^{{ib}} } \\ $$$$\mathrm{tan}{ib}\:\:=\:{i}\frac{{e}^{{b}} −{e}^{−{ib}} }{{e}^{{ib}} +{e}^{−{ib}} }\:=\:{i}\frac{\mathrm{sinh}{b}}{\mathrm{cosh}{b}}\:=\:{i}\mathrm{tanh}{b} \\ $$$$ \\ $$$${x}+{iy}\:=\:\frac{\mathrm{tan}{a}+{i}\mathrm{tanh}{b}}{\mathrm{1}−{i}\mathrm{tan}{a}\mathrm{tanh}{b}} \\ $$$${x}+{iy}\:=\:\frac{\left(\mathrm{tan}{a}+{i}\mathrm{tanh}{b}\right)\left(\mathrm{1}+{i}\mathrm{tan}{a}\mathrm{tanh}{b}\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {a}\mathrm{tanh}^{\mathrm{2}} {b}} \\ $$$${x}+{iy}\:=\:\frac{\mathrm{tan}{a}\left(\mathrm{1}−\mathrm{tanh}^{\mathrm{2}} {b}\right)+{i}\mathrm{tanh}{b}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {a}\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {a}\mathrm{tanh}^{\mathrm{2}} {b}} \\ $$$$ \\ $$$$\frac{{y}}{{x}}\:=\:\frac{\mathrm{tanh}{b}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {a}\right)}{\mathrm{tan}{a}\left(\mathrm{1}−\mathrm{tanh}^{\mathrm{2}} {b}\right)} \\ $$$$\frac{{y}}{{x}}\:=\:\frac{\mathrm{tanh}{b}\mathrm{cosh}^{\mathrm{2}} {b}}{\mathrm{tan}{a}\mathrm{cos}^{\mathrm{2}} {a}}\:=\:\frac{\mathrm{sinh}{b}\mathrm{cosh}{b}}{\mathrm{sin}{a}\mathrm{cos}{a}} \\ $$$$\frac{{y}}{{x}}\:=\:\frac{\mathrm{tanh}{b}\mathrm{cosh}^{\mathrm{2}} {b}}{\mathrm{tan}{a}\mathrm{cos}^{\mathrm{2}} {a}}\:=\:\frac{\mathrm{sinh2}{b}}{\mathrm{sin2}{a}}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\frac{\mathrm{tan}^{\mathrm{2}} {a}}{\mathrm{cosh}^{\mathrm{4}} {b}}+\frac{\mathrm{tanh}^{\mathrm{2}} {b}}{\mathrm{cos}^{\mathrm{4}} {a}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\frac{\mathrm{sin}^{\mathrm{2}} {a}\mathrm{cos}^{\mathrm{2}} {a}+\mathrm{sinh}^{\mathrm{2}} {b}\mathrm{cosh}^{\mathrm{2}} {b}}{\mathrm{cos}^{\mathrm{4}} {a}\mathrm{cosh}^{\mathrm{4}} {b}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{2}{a}+\mathrm{sinh}^{\mathrm{2}} \mathrm{2}{b}}{\mathrm{4cos}^{\mathrm{4}} {a}\mathrm{cosh}^{\mathrm{4}} {b}}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{work}\:\mathrm{in}\:\mathrm{progress}... \\ $$$$ \\ $$

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