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Knowing $w=\arccos(z)=-i\log(z+(z^2-1)^{1/2})$.

Solve $\cos(z)=-i$ in the form $a+bi$.

Can someone help me solve this? I have been trying but I don't have any solutions so I don't know how I am supposed to do this. Thanks in advance.

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    it is $$\pi /2-i\ln \left( 1+\sqrt {2} \right) $$ – 2017-01-02
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    Can you show how you got that? I have $z=-ilog(i(-1+-2^{1/2})$ – 2017-01-02
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    i have given you a hint below – 2017-01-02
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    Follow Travis's method in a similar problem [here](http://math.stackexchange.com/questions/1242873/solving-cos-z-i-for-z). – 2017-01-02
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    Vasco: Please explain how the "answer" you accepted helps you to go further than the formula in your comment above and in particular, to answer the question you asked? – 2017-01-02
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    @Did I accepted that answer because that is the way i remember learning it in class and i believe that is the way i am supposed to do it. I am still trying to figure out how to go past what i wrote in the other comment. – 2017-01-02
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    IOW, the "answer" merely repeats things you already knew and it does not help you answer your question. The logic of immediately accepting it (did you do that merely 16 minutes after said "answer" was posted?) escapes me, I am afraid. And, say, did you even *read* the other answer you received? – 2017-01-02

2 Answers 2

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HINT: write your equation in the form $$\frac{1}{2}(e^{iz}+e^{-iz})=i$$

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    That is what i did and now i have $z=-ilog(i(-1+-(2^{1/2}))$ – 2017-01-02
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    @VascoMartins - Are you sure your answer isn't the same as his? Consider the $\log$ laws. – 2017-01-02
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    as it stands, this answer is useless :/ – 2017-01-02
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Set $z=x+iy$, so $$ \cos z=\cos x\cos(iy)-\sin x\sin(iy)=\cos x\cosh y+i\sin x\sinh y $$ Thus $$ \begin{cases} \cos x\cosh y=0 \\[4px] \sin x\sinh y=-1 \end{cases} $$