How can I proceed to solve $$\cos(z)=\frac 34+\frac i4$$ I'm not very good in complex variable... But I know the definition of the complex functions, can you help me?
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3$\arccos (\frac{3}{4} + \frac{i}{4})$ ? – John Kontol Mar 30 '17 at 13:25
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1Ok, but I don't know how to compute the arccos in complex, I mean the non-inverse trigonometric functions. – Ragnar1204 Mar 30 '17 at 13:26
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3Express $\cos z$ in terms of the real and imaginary parts of $z$ (the addition formula can help). – Mar 30 '17 at 13:30
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I've got that $(e^{-y}-e^{y})cos(x)=\frac{3}{4}$ and $(e^{-y}-e^{y})sin(x)=\frac{1}{4}$... then? – Ragnar1204 Mar 30 '17 at 13:33
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1Divide those equations to find $\tan (x)$ – Sebastian Schulz Mar 30 '17 at 13:39
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You're right.. thank you very much!!!! – Ragnar1204 Mar 30 '17 at 13:40
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As you indicated, you know the definitions of complex functions, hence I am sure you know the definition of the complex cosine to be $cosz=\frac{e^{iz}+e^{-iz}}{2}$ and so using a general approach for your equation, we wish to solve $\frac{e^{iz}+e^{-iz}}{2}=\frac{3}{4}+\frac{1}{4}i$ which simplifies to the quadratic equation $e^{iz}+e^{-iz}=\frac{3}{2}+\frac{1}{2}i$. Now multiplying through by $e^{iz}$ and setting $e^{iz}=t$, we arrive at the quadratic equation $t^2-\frac{3}{2}t-\frac{1}{2}it+1=0$. The best way to solve a quadratic equation with complex coefficients is through Completing the Square. So after bringing over the $1$ to the right side, we get (please verify!): $$(t-\frac{3}{4}-\frac{1}{4}i)^2=-1+(-\frac{3}{4}-\frac{1}{4}i)^2$$ Working out the right hand side, this simplifies to $-\frac{1}{2}+\frac{3}{8}i$. Using deMoivres theorem (given $r=\frac{5}{8}$ and argument $\theta=143.13010.....$do not round!) you find $$(t-\frac{3}{4}-\frac{1}{4}i)=\pm(\frac{1}{4}+\frac{3}{4}i)$$ from which you can solve $t$ to be $1+i$ and $\frac{1}{2}-\frac{1}{2}i$. Side note, it is very cool that these numbers come out "nice". So now you have to solve $e^{iz}=1+i$ and $e^{iz}=\frac{1}{2}-\frac{1}{2}i$ for $z$ which can be done through the complex logarithm. That I'll leave up to you.
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