MHB What is the ratio of complex numbers in the form of a question?

AI Thread Summary
The ratio $\frac{e^{i\sqrt{x}}-1}{e^{i\sqrt{x}}+1}$ simplifies to $i\tan\left(\frac{\sqrt{x}}{2}\right)$. The discussion highlights the use of Euler's formulas for sine and cosine to derive this result. Participants express concern about understanding the derivation for exam purposes. The final consensus emphasizes the importance of remembering this formula. This mathematical relationship is crucial for solving related complex number problems.
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What's the ratio $\displaystyle \frac{e^{i\sqrt{x}}-1}{e^{i\sqrt{x}}+1}$ equal to? I can't work it out to anything I recognize. :confused:

The answer is $\displaystyle i\tan(\frac{1}{2}\sqrt{x})$. I suppose I could work backwards from the answer, but I won't have the answer in the exam.
 
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See how far you can get using $$\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$$ and $$\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$$.
 
Guest said:
What's the ratio $\displaystyle \frac{e^{i\sqrt{x}}-1}{e^{i\sqrt{x}}+1}$ equal to? I can't work it out to anything I recognize. :confused:

The answer is $\displaystyle i\tan(\frac{1}{2}\sqrt{x})$. I suppose I could work backwards from the answer, but I won't have the answer in the exam.

$\displaystyle \begin{align*} \frac{\mathrm{e}^{\mathrm{i}\,\sqrt{x}}-1}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} &= \frac{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1 - 2}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} \\ &= 1 - \frac{2}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} \\ &= 1 - \frac{2}{1 + \cos{ \left( \sqrt{x} \right) } + \mathrm{i} \sin{\left( \sqrt{x} \right) }} \\ &= 1 - \frac{2\,\left[ 1 + \cos{ \left( \sqrt{x} \right) } - \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] }{\left[ 1 + \cos{\left( \sqrt{x} \right) } + \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] \left[ 1 + \cos{ \left( \sqrt{x} \right) } - \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right)
} - 2\,\mathrm{i} \sin{ \left( \sqrt{x} \right) } }{ \left[ 1 + \cos{ \left( \sqrt{x} \right) } \right] ^2 + \sin^2{ \left( \sqrt{x} \right) } } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right) } - 2\,\mathrm{i}\sin{\left( \sqrt{x} \right) }}{1 + 2\cos{ \left( \sqrt{x} \right) } + \cos^2{ \left( \sqrt{x} \right) } + \sin^2{ \left( \sqrt{x} \right) } } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right) } - 2\,\mathrm{i}\sin{\left( \sqrt{x} \right) }}{2 + 2\cos{\left( \sqrt{x} \right) }} \\ &= 1 - \left[ 1 - \frac{2\,\mathrm{i}\sin{\left( \sqrt{x} \right) } }{2\,\left[ 1 + \cos{\left( \sqrt{x} \right) } \right] } \right] \\ &= \mathrm{i}\,\left[ \frac{\sin{\left( \sqrt{x} \right) }}{1 + \cos{ \left( \sqrt{x} \right) } } \right] \\ &= \mathrm{i}\,\left[ \frac{2\sin{\left( \frac{\sqrt{x}}{2} \right) } \cos{\left( \frac{\sqrt{x}}{2} \right) }}{1 + 2\cos^2{\left( \frac{\sqrt{x}}{2} \right) } - 1 } \right] \\ &= \mathrm{i}\,\left[ \frac{2\sin{\left( \frac{\sqrt{x}}{2} \right) }\cos{\left( \frac{\sqrt{x}}{2}\right) }}{2\cos^2{\left( \frac{\sqrt{x}}{2} \right) }} \right] \\ &= \mathrm{i}\tan{ \left( \frac{\sqrt{x}}{2} \right) } \end{align*}$
 
$$\dfrac{e^{i\sqrt x}-1}{e^{i\sqrt x}+1}=\dfrac{e^{i\sqrt x/2}-e^{-i\sqrt x/2}}{e^{i\sqrt x/2}+e^{-i\sqrt x/2}}=i\dfrac{\sin\dfrac{\sqrt x}{2}}{\cos\dfrac{\sqrt x}{2}}=i\tan\dfrac{\sqrt x}{2}$$
 
Thanks, guys. I appreciate this. :D

greg1313 said:
$$\dfrac{e^{i\sqrt x}-1}{e^{i\sqrt x}+1}=\dfrac{e^{i\sqrt x/2}-e^{-i\sqrt x/2}}{e^{i\sqrt x/2}+e^{-i\sqrt x/2}}=i\dfrac{\sin\dfrac{\sqrt x}{2}}{\cos\dfrac{\sqrt x}{2}}=i\tan\dfrac{\sqrt x}{2}$$
I must remember this one for my exam. Sweet and short!
 
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