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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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