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Question Number 88594 by mr W last updated on 11/Apr/20
solve for x∈C  cos (x)=a+bi
$${solve}\:{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{cos}\:\left({x}\right)={a}+{bi} \\ $$
Commented by abdomathmax last updated on 11/Apr/20
⇒ch(ix) =a+bi ⇒((e^(ix)  +e^(−ix) )/2)=a+bi ⇒  e^(ix)  +e^(−ix)  =2a+2bi  let e^(ix) =z ⇒ix =ln(z) ⇒  x =−iln(z)  (e)→z+z^(−1)  =2a+2bi ⇒z^2  +1 =(2a+2bi)z ⇒  z^2 −2(a+bi)z +1 =0  Δ^′ =(a+bi)^2 −1 =a^2  +2abi−b^2 −1  =a^2 −b^2 −1 +2abi =(√((a^2 −b^2 −1)^2  +4a^2 b^2 ))e^(iarctan(((2ab)/(a^2 −b^2 −1))))   ⇒z_1 =((a+bi +((a^2 −b^2 −1)^2  +4a^2 b^2 )^(1/4)  e^((i/2)arcran(((2ab)/(a^2 −b^2 −1)))) )/1)  z_2 =a+bi+{(a^2 −b^2 −1)^2 +4a^2 b^2 }^(1/4)  e^((i/2) arctan(((2ab)/(a^2 −b^2 −1))))   ⇒x =−iln( a+bi +^− {(a^2 −b^2 −1)^2 +4a^2 b^2 }^(1/4)  e^((i/2)arctan(((2ab)/(a^2 −b^2 −1))))
$$\Rightarrow{ch}\left({ix}\right)\:={a}+{bi}\:\Rightarrow\frac{{e}^{{ix}} \:+{e}^{−{ix}} }{\mathrm{2}}={a}+{bi}\:\Rightarrow \\ $$$${e}^{{ix}} \:+{e}^{−{ix}} \:=\mathrm{2}{a}+\mathrm{2}{bi}\:\:{let}\:{e}^{{ix}} ={z}\:\Rightarrow{ix}\:={ln}\left({z}\right)\:\Rightarrow \\ $$$${x}\:=−{iln}\left({z}\right) \\ $$$$\left({e}\right)\rightarrow{z}+{z}^{−\mathrm{1}} \:=\mathrm{2}{a}+\mathrm{2}{bi}\:\Rightarrow{z}^{\mathrm{2}} \:+\mathrm{1}\:=\left(\mathrm{2}{a}+\mathrm{2}{bi}\right){z}\:\Rightarrow \\ $$$${z}^{\mathrm{2}} −\mathrm{2}\left({a}+{bi}\right){z}\:+\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta^{'} =\left({a}+{bi}\right)^{\mathrm{2}} −\mathrm{1}\:={a}^{\mathrm{2}} \:+\mathrm{2}{abi}−{b}^{\mathrm{2}} −\mathrm{1} \\ $$$$={a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}\:+\mathrm{2}{abi}\:=\sqrt{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{e}^{{iarctan}\left(\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$\Rightarrow{z}_{\mathrm{1}} =\frac{{a}+{bi}\:+\left(\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arcran}\left(\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}}\right)} }{\mathrm{1}} \\ $$$${z}_{\mathrm{2}} ={a}+{bi}+\left\{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right\}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$\Rightarrow{x}\:=−{iln}\left(\:{a}+{bi}\:\overset{−} {+}\left\{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right\}^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{2}{ab}}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{1}}\right)} \right. \\ $$$$ \\ $$
Commented by mr W last updated on 11/Apr/20
thank you sir! can you please put it into  x+yi form?
$${thank}\:{you}\:{sir}!\:{can}\:{you}\:{please}\:{put}\:{it}\:{into} \\ $$$${x}+{yi}\:{form}? \\ $$
Commented by abdomathmax last updated on 11/Apr/20
you are welcome you can sir use  u+iv =(√(u^2  +v^2 )) e^(isrctan((v/u)))  ⇒  ln(u+iv) =(1/2)ln(u^2  +v^2 )+iarctan((v/u)) ...now its  eazy with that...
$${you}\:{are}\:{welcome}\:{you}\:{can}\:{sir}\:{use} \\ $$$${u}+{iv}\:=\sqrt{{u}^{\mathrm{2}} \:+{v}^{\mathrm{2}} }\:{e}^{{isrctan}\left(\frac{{v}}{{u}}\right)} \:\Rightarrow \\ $$$${ln}\left({u}+{iv}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} \:+{v}^{\mathrm{2}} \right)+{iarctan}\left(\frac{{v}}{{u}}\right)\:…{now}\:{its} \\ $$$${eazy}\:{with}\:{that}… \\ $$

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