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Find all complex solutions or $\sin(z)=0.75i$

I started with $$\frac{e^{iz}-e^{-iz}}{2i}=0.75i$$

$${e^{iz}-e^{-iz}}=-1.5$$

$${e^{iz}-e^{-iz}}=e^{ln(-1.5)}$$

But it is keeping get complicated

Did
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newhere
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    Hint: let $t=e^{iz}$. – Ixion Aug 17 '17 at 12:50
  • so we are left with $e^{iz}=-2$ and $e^{iz}=0.5$ – newhere Aug 17 '17 at 13:07
  • Yes, $e^{iz}=-2$ OR $e^{iz}=0.5$. Now, since $z\in\mathbb{C}$ then $z=a+ib\ \ \ \mbox{with} \ \ \ a,b\in\mathbb{R}$ so $$e^{iz}=e^{-b+ai}=e^{-b}(\cos(a)+i\sin(a))$$, hence $$Re(e^{iz})=e^{-b}\cos(a)\mbox{ and } Im(e^{iz})=e^{-b}\sin(a)$$ so... – Ixion Aug 17 '17 at 13:13
  • so we have $e^{-b}=-2$ and $a=2\pi k$ – newhere Aug 17 '17 at 13:19
  • @Ixion Simpler than $\Re$ and $\Im$, use that $e^{ia}e^{-b}=w$ if and only if $e^{-b}=|w|$ and $a\in\mathrm{Arg}(w)+2\pi\mathbb Z$. For $w=-2$, $|w|=2$ and $\mathrm{Arg}(w)=\pi$ hence $b=-\ln2$ and $a\in\pi+2\pi\mathbb Z$, that is, $z=\pi+2k\pi-i\ln2$ with $k\in\mathbb Z$. Similarly for $w=\frac12$. – Did Aug 17 '17 at 14:15
  • @Did in $0.5$ we have $e^{-b}=0.5$ so we have $-iln(0.5)$? – newhere Aug 17 '17 at 14:19
  • What? $ $ $ $ $ $ – Did Aug 17 '17 at 15:07
  • @Did Sorry, I managed – newhere Aug 17 '17 at 15:13

2 Answers2

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${e^{iz}-e^{-iz}}=-1.5$

Set $e^{iz}=t$ so the equation becomes

$t-\dfrac{1}{t}+1.5==0$

$t_1=-2;\;t=\dfrac{1}{2}$

so $e^{iz}=-2$ which leads to $z= (2 c_1+1)\pi-i \log 2,\forall c_1\in\mathbb{Z}$

and $e^{iz}=\dfrac{1}{2}$ which gives $z=2 \pi c_2+i \log 2,\forall c_2\in\mathbb{Z}$

Hope this is useful

Edit

$e^{i z}=\dfrac{1}{2}$

$e^{i z}=e^{\log \frac{1}{2}}$

$iz=\log \frac{1}{2}+2k\pi i,\forall k\in\mathbb{Z}$

$z=\dfrac{1}{i}\,(-\log 2+2k\pi i)$

$z=-i(-\log 2+2k\pi i)$

$z=-2k \pi+i\log 2,\;\forall k\in\mathbb{Z}$

can be written also as $z=2k \pi+i\log 2,\;\forall k\in\mathbb{Z}$

Raffaele
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Another way to look at this: Start with

$$\sin z=\frac{e^{iz}-e^{-iz}}{2i}=0.75i$$

Let $w=iz$, so that we get

$$\sinh w=-\frac{3}{4}\\ w=\sinh^{-1}\frac{3}{4}=\ln \frac{1}{2}=-\ln 2$$

and finally,

$$z=i\ln 2$$

Cye Waldman
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