I-need-the-sol-plz-find-the-imaginary-and-real-parts-of-log-sin-a-ib-

Question Number 73913 by arkanmath7@gmail.com last updated on 16/Nov/19

$${I}\:{need}\:{the}\:{sol}.\:{plz} \\ $$$${find}\:{the}\:{imaginary}\:{and}\:{real}\:{parts}\:{of} \\ $$$${log}\:{sin}\left({a}+{ib}\right)? \\ $$

Answered by Tanmay chaudhury last updated on 16/Nov/19

Log(sinacosib+cosasinib) Log(sina.coshb+cosa.isinhb) Rcosα=sina.coshb Rsinα=cosa.sinhb Log(Rcosα+iRsinα) Log[Re^(iα) ]⇛Log[Re^(i(2nπ+α)) ] Log[Re^(i(2nπ+α)) ]=logR+i(2nπ+α) R=[(sina.coshb)^2 +(cosa.sinhb)^2 ]^(1/2) tanα=((cosa.sinhb)/(sina.coshb))⇛α=tan^(−1) (((cosa.sinhb)/(sina.coshb)))

$${Log}\left({sinacosib}+{cosasinib}\right) \\ $$$${Log}\left({sina}.{coshb}+\mathrm{c}{osa}.{isinhb}\right) \\ $$$${Rcos}\alpha={sina}.{coshb} \\ $$$${Rsin}\alpha={cosa}.{sinhb} \\ $$$${Log}\left({Rcos}\alpha+{iRsin}\alpha\right) \\ $$$${Log}\left[{Re}^{{i}\alpha} \right]\Rrightarrow{Log}\left[{Re}^{{i}\left(\mathrm{2}{n}\pi+\alpha\right)} \right] \\ $$$${Log}\left[{Re}^{{i}\left(\mathrm{2}{n}\pi+\alpha\right)} \right]={logR}+{i}\left(\mathrm{2}{n}\pi+\alpha\right) \\ $$$${R}=\left[\left({sina}.{coshb}\right)^{\mathrm{2}} +\left({cosa}.{sinhb}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${tan}\alpha=\frac{{cosa}.{sinhb}}{{sina}.{coshb}}\Rrightarrow\alpha={tan}^{−\mathrm{1}} \left(\frac{{cosa}.{sinhb}}{{sina}.{coshb}}\right) \\ $$

Commented by arkanmath7@gmail.com last updated on 16/Nov/19

$${thnx} \\ $$

Answered by Smail last updated on 16/Nov/19

ln(sin(a+ib))=x+iy sin(a+ib)=e^x e^(iy) sin(a)cos(ib)+sin(ib)cos(a)=e^x e^(iy) sin(a)cosh(b)+isinh(b)cos(a)=e^x e^(iy) ((sinh(b)cos(a))/(sin(a)cosh(b)))=tan(y) y=tan^(−1) (tanh(b)cot(a))) e^x =(√(sin^2 (a)cosh^2 (b)+sinh^2 (b)cos^2 (a))) x=(1/2)ln(sin^2 (a)cosh^2 (b)+sinh^2 (b)cos^2 (a)) sin(a+ib)=(1/2)ln(sin^2 (a)cosh^2 (b)+sinh^2 (b)cos^2 (a))+itan^(−1) (tanh(b)cot(a))

$${ln}\left({sin}\left({a}+{ib}\right)\right)={x}+{iy} \\ $$$${sin}\left({a}+{ib}\right)={e}^{{x}} {e}^{{iy}} \\ $$$${sin}\left({a}\right){cos}\left({ib}\right)+{sin}\left({ib}\right){cos}\left({a}\right)={e}^{{x}} {e}^{{iy}} \\ $$$${sin}\left({a}\right){cosh}\left({b}\right)+{isinh}\left({b}\right){cos}\left({a}\right)={e}^{{x}} {e}^{{iy}} \\ $$$$\frac{{sinh}\left({b}\right){cos}\left({a}\right)}{{sin}\left({a}\right){cosh}\left({b}\right)}={tan}\left({y}\right) \\ $$$$\left.{y}={tan}^{−\mathrm{1}} \left({tanh}\left({b}\right){cot}\left({a}\right)\right)\right) \\ $$$${e}^{{x}} =\sqrt{{sin}^{\mathrm{2}} \left({a}\right){cosh}^{\mathrm{2}} \left({b}\right)+{sinh}^{\mathrm{2}} \left({b}\right){cos}^{\mathrm{2}} \left({a}\right)} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}^{\mathrm{2}} \left({a}\right){cosh}^{\mathrm{2}} \left({b}\right)+{sinh}^{\mathrm{2}} \left({b}\right){cos}^{\mathrm{2}} \left({a}\right)\right) \\ $$$${sin}\left({a}+{ib}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}^{\mathrm{2}} \left({a}\right){cosh}^{\mathrm{2}} \left({b}\right)+{sinh}^{\mathrm{2}} \left({b}\right){cos}^{\mathrm{2}} \left({a}\right)\right)+{itan}^{−\mathrm{1}} \left({tanh}\left({b}\right){cot}\left({a}\right)\right) \\ $$

Commented by Smail last updated on 16/Nov/19

ln(sin(a+ib))=(1/2)ln(sin^2 (a)cosh^2 (b)+sinh^2 (b)cos^2 (a))+itan^(−1) (tanh(b)cot(a))

$${ln}\left({sin}\left({a}+{ib}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}^{\mathrm{2}} \left({a}\right){cosh}^{\mathrm{2}} \left({b}\right)+{sinh}^{\mathrm{2}} \left({b}\right){cos}^{\mathrm{2}} \left({a}\right)\right)+{itan}^{−\mathrm{1}} \left({tanh}\left({b}\right){cot}\left({a}\right)\right) \\ $$$$ \\ $$

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