6

For this question I tried to use this identity:

$$\sin(z)= \frac{e^{iz}-e^{-iz}}{2i}$$

and let $\sin(z)= \dfrac{(e^{iz}-e^{-iz})}{2i}=1-i$, then $e^{iz}-e^{-iz}-(2i+2)=0$;

multiply both sides of this equation by $e^{iz}$,and let $w=e^{iz}$,

$w^2-(2i+2)w-1=0$,

I got

$w=(1+i)\pm\sqrt{1+2i}$,

what can I do next?

Nosrati
  • 29,995
Jone Will
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  • Write $z =x+iy$ in $w=e^{iz}$ $\rightarrow$ $e^{-y} = |w| $ – john doe Nov 05 '17 at 20:22
  • $e^{a+ib} = e^ae^{ib}$ – amsmath Nov 05 '17 at 20:23
  • i have found this here $$\arcsin \left( 1/2,\sqrt {5}-1/2 \right) -i\ln \left( 1/2,\sqrt {5} +1/2+\sqrt { \left( 1/2,\sqrt {5}+1/2 \right) ^{2}-1} \right) $$ – Dr. Sonnhard Graubner Nov 05 '17 at 20:29
  • You simply need to solve $$u^2=1+2i$$ then $$e^{iz}=1+i+u$$ To find $u$, use $u=\xi+i\eta$ and find $(\xi,\eta)$ by solving $$\xi^2-\eta^2=\xi\eta=1$$ To deduce $z$, use $z=x+iy$ then $$e^x=|1+i+u|\qquad e^{iy}=\frac{1+i+u}{|1+i+u|}$$ – Did Nov 05 '17 at 21:26

2 Answers2

1

You can use one of the complex root formulas, one is $$ \sqrt z=\sqrt{|z|}\frac{|z|+z}{||z|+z|} $$ which would give here $$ \sqrt{1+2i}=\sqrt[4]{5}\frac{\sqrt5+1+2i}{\sqrt{10+2\sqrt5}}=\frac{(1+\sqrt5)+2i}{\sqrt{2+2\sqrt5}} $$

Lutz Lehmann
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0

Apply the identity $\arcsin (w) = -i \ln \left(iw + \sqrt{1 - w^2}\right)$ to obtain the solutions as follows,

\begin{align} z&= \arcsin (1-i ) \\ &= \arcsin (\sqrt2e^{-i\frac\pi4}) \\ & = -i \ln \left(\sqrt2e^{i\frac{\pi}4} \pm \sqrt[4]5e^{\frac{i }2 \arcsin\frac2{\sqrt5} }\right) \\ &= -i \ln \left(1\pm \phi^{1/2}+ i \left( 1\pm \ \phi^{-1/2}\right) \right), \>\>\> \phi=\frac{\sqrt5+1}2\\ &= -i \ln \left(r_{\pm}e^{\pm i \theta+i2\pi k}\right) ,\>\>\>r_{\pm}^2 =\sqrt5+2\pm2\sqrt{\sqrt5+2},\>\tan\theta =\phi^{-1/2}\\ &= 2\pi k\pm \theta -i \ln r_{\pm} \\ &=2\pi k\pm \arctan\sqrt{\frac{\sqrt5-1}2}-\frac i2\ln \left(\sqrt5+2\pm 2\sqrt{\sqrt5+2}\right) \end{align}

Quanto
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