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Question Number 50849 by peter frank last updated on 21/Dec/18

solve for z in the form x+iy if tanz=0.5

$$\mathrm{solve}\:{for}\:{z}\:\:{in}\:{the}\:{form}\:\:{x}+{iy}\: \\ $$$${if}\:{tanz}=\mathrm{0}.\mathrm{5}\: \\ $$

Answered by ajfour last updated on 21/Dec/18

e^(iz) = cos z+isin z e^(−iz) = cos z−isin z tan z= ((e^(iz) −e^(−iz) )/(i(e^(iz) +e^(−iz) ))) let e^(iz) = e^(−y+ix) = e^(−y) (cos x+isin x) ⇒ e^(−iz) = e^(y−ix) = e^y (cos x−isin x) ⇒ tan z = (((e^(−y) −e^y )cos x+i(e^(−y) +e^y )sin x)/(i[(e^(−y) +e^y )cos x+i(e^(−y) −e^y )sin x])) let e^(−y) +e^y = u , e^(−y) −e^y = v tan z = (((−ivcos x+usin x)(ucos x−ivsin x))/(u^2 cos^2 x+v^2 sin^2 x)) tan z = ((−iuv+(u^2 −v^2 )sin 2x)/(2(u^2 cos^2 x+v^2 sin^2 x))) if tan z = (1/2) , ⇒ uv = 0 or e^(−2y) =e^(2y) ⇒ y = 0 ⇒ u = 2 , v = 0 (((u^2 −v^2 )sin 2x)/(2(u^2 cos^2 x+v^2 sin^2 x))) = (1/2) ⇒ tan x = (1/2) ⇒ z = x = tan^(−1) (1/2)+nπ .

$${e}^{{iz}} \:=\:\mathrm{cos}\:{z}+{i}\mathrm{sin}\:{z} \\ $$$${e}^{−{iz}} \:=\:\mathrm{cos}\:{z}−{i}\mathrm{sin}\:{z} \\ $$$$\mathrm{tan}\:{z}=\:\frac{{e}^{{iz}} −{e}^{−{iz}} }{{i}\left({e}^{{iz}} +{e}^{−{iz}} \right)}\: \\ $$$${let}\:\:{e}^{{iz}} \:=\:{e}^{−{y}+{ix}} \:=\:{e}^{−{y}} \left(\mathrm{cos}\:{x}+{i}\mathrm{sin}\:{x}\right) \\ $$$$\Rightarrow\:\:\:{e}^{−{iz}} \:=\:{e}^{{y}−{ix}} =\:{e}^{{y}} \left(\mathrm{cos}\:{x}−{i}\mathrm{sin}\:{x}\right) \\ $$$$\Rightarrow\:\mathrm{tan}\:{z}\:=\:\frac{\left({e}^{−{y}} −{e}^{{y}} \right)\mathrm{cos}\:{x}+{i}\left({e}^{−{y}} +{e}^{{y}} \right)\mathrm{sin}\:{x}}{{i}\left[\left({e}^{−{y}} +{e}^{{y}} \right)\mathrm{cos}\:{x}+{i}\left({e}^{−{y}} −{e}^{{y}} \right)\mathrm{sin}\:{x}\right]} \\ $$$${let}\:\:{e}^{−{y}} +{e}^{{y}} \:=\:{u}\:\:,\:\:{e}^{−{y}} −{e}^{{y}} \:=\:{v} \\ $$$$\mathrm{tan}\:{z}\:=\:\frac{\left(−{iv}\mathrm{cos}\:{x}+{u}\mathrm{sin}\:{x}\right)\left({u}\mathrm{cos}\:{x}−{iv}\mathrm{sin}\:{x}\right)}{{u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}} \\ $$$$\:\mathrm{tan}\:{z}\:=\:\frac{−{iuv}+\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}\left({u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}\right)} \\ $$$${if}\:\:\mathrm{tan}\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:,\:\Rightarrow \\ $$$$\:\:\:\:{uv}\:=\:\mathrm{0}\:\:\:{or}\:\:\:{e}^{−\mathrm{2}{y}} ={e}^{\mathrm{2}{y}} \:\:\: \\ $$$$\Rightarrow\:\:\:\:\:{y}\:=\:\mathrm{0}\:\:\:\: \\ $$$$\Rightarrow\:\:\:{u}\:=\:\mathrm{2}\:,\:\:{v}\:=\:\mathrm{0} \\ $$$$\:\frac{\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}\left({u}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} {x}+{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} {x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\: \\ $$$$\Rightarrow\:\:\:\boldsymbol{{z}}\:=\:{x}\:=\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}+{n}\pi\:\:\:\:. \\ $$

Commented by peter frank last updated on 21/Dec/18

ans [ z=((πn)/2)+26.56+0.00000119i]

$${ans}\:\left[\:\:{z}=\frac{\pi\mathrm{n}}{\mathrm{2}}+\mathrm{26}.\mathrm{56}+\mathrm{0}.\mathrm{00000119}{i}\right] \\ $$

Commented by MJS last updated on 21/Dec/18

sorry but this is wrong. possibly the calculator approximates in a crazy way. tan (a+bi) =(1/2) has only real solutions ⇒ b=0

$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{possibly}\:\mathrm{the}\:\mathrm{calculator} \\ $$$$\mathrm{approximates}\:\mathrm{in}\:\mathrm{a}\:\mathrm{crazy}\:\mathrm{way}. \\ $$$$\mathrm{tan}\:\left({a}+{b}\mathrm{i}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{has}\:\mathrm{only}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:{b}=\mathrm{0} \\ $$

Commented by peter frank last updated on 22/Dec/18

thank you both sirs

$${thank}\:{you}\:{both}\:{sirs} \\ $$

Answered by MJS last updated on 21/Dec/18

tan (x+iy) =(1/2); ((x),(y) )∈R^2 x+iy=nπ+arctan (1/2) ; n∈Z ⇒ y=0; x=nπ+arctan (1/2); n∈Z

$$\mathrm{tan}\:\left({x}+\mathrm{i}{y}\right)\:=\frac{\mathrm{1}}{\mathrm{2}};\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\in\mathbb{R}^{\mathrm{2}} \\ $$$${x}+\mathrm{i}{y}={n}\pi+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:{n}\in\mathbb{Z} \\ $$$$\Rightarrow\:{y}=\mathrm{0};\:{x}={n}\pi+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}};\:{n}\in\mathbb{Z} \\ $$

Answered by peter frank last updated on 21/Dec/18

tanh (iz)=itanz (1/2)=((tanh (iz))/i) (1/2)=((e^(iz) −e^(−iz) )/(i(e^(iz) +e^(−iz) ))) multiply by e^(iz) and simplify e^(2iz) =(3/5)+((4i)/5) z=x+iy e^(2ix) .e^(−2y) =(3/5)+((4i)/5) from Eurel′s form of complex no e^(2xi) =cos 2x+isin 2x e^(−2y) (cos 2x+isin 2x)=(3/4)+((4i)/5) compare real and Immarginary e^(−2y) cos 2x=(3/5)....(i) e^(−2y) sin 2x=(4/5)....(ii) divide ii by i tan2x=(4/3) tan2x=tan (53.13010235)≈tan 53.13 2x=πn+53.13^° x=((πn)/2)+26.565 n=0,1,2,3 .... from e^(−2y) cos 2x=(3/5)....(i) e^(−2y) =((0.6)/(cos 2x)) −2y=ln(((0.6)/(cos 2x))) y=(1/2)ln(((0.6)/(cos 2x))) z=x+iy z=((πn)/2)+26.565−(1/2)ln(((0.6)/(cos 2x))) ......

$$\:\mathrm{tanh}\:\left({iz}\right)={itanz} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{tanh}\:\left({iz}\right)}{{i}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\frac{{e}^{{iz}} −{e}^{−{iz}} }{{i}\left({e}^{{iz}} +{e}^{−{iz}} \right)} \\ $$$${multiply}\:{by}\:{e}^{{iz}} \:{and}\:{simplify} \\ $$$${e}^{\mathrm{2}{iz}} =\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{4}{i}}{\mathrm{5}} \\ $$$${z}={x}+{iy} \\ $$$${e}^{\mathrm{2}{ix}} .{e}^{−\mathrm{2}{y}} =\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{4}{i}}{\mathrm{5}} \\ $$$${from}\:{Eurel}'{s}\:{form}\:{of}\: \\ $$$${complex}\:{no} \\ $$$${e}^{\mathrm{2}{xi}} =\mathrm{cos}\:\mathrm{2}{x}+{i}\mathrm{sin}\:\mathrm{2}{x} \\ $$$${e}^{−\mathrm{2}{y}} \left(\mathrm{cos}\:\mathrm{2}{x}+{i}\mathrm{sin}\:\mathrm{2}{x}\right)=\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{4}{i}}{\mathrm{5}} \\ $$$${compare}\:{real}\:{and}\:{Immarginary} \\ $$$${e}^{−\mathrm{2}{y}} \mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{3}}{\mathrm{5}}....\left({i}\right) \\ $$$${e}^{−\mathrm{2}{y}} \mathrm{sin}\:\:\mathrm{2}{x}=\frac{\mathrm{4}}{\mathrm{5}}....\left({ii}\right) \\ $$$${divide}\:{ii}\:{by}\:{i} \\ $$$${tan}\mathrm{2}{x}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${tan}\mathrm{2}{x}=\mathrm{tan}\:\left(\mathrm{53}.\mathrm{13010235}\right)\approx\mathrm{tan}\:\mathrm{53}.\mathrm{13} \\ $$$$\mathrm{2}{x}=\pi{n}+\mathrm{53}.\mathrm{13}^{°} \\ $$$${x}=\frac{\pi{n}}{\mathrm{2}}+\mathrm{26}.\mathrm{565} \\ $$$${n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$.... \\ $$$${from} \\ $$$${e}^{−\mathrm{2}{y}} \mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{3}}{\mathrm{5}}....\left({i}\right) \\ $$$${e}^{−\mathrm{2}{y}} =\frac{\mathrm{0}.\mathrm{6}}{\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$−\mathrm{2}{y}={ln}\left(\frac{\mathrm{0}.\mathrm{6}}{\mathrm{cos}\:\mathrm{2}{x}}\right) \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{0}.\mathrm{6}}{\mathrm{cos}\:\mathrm{2}{x}}\right) \\ $$$${z}={x}+{iy} \\ $$$${z}=\frac{\pi{n}}{\mathrm{2}}+\mathrm{26}.\mathrm{565}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{0}.\mathrm{6}}{\mathrm{cos}\:\mathrm{2}{x}}\right) \\ $$$$...... \\ $$

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