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

[Guest2]

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.

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.

 

[Greg]

See how far you can get using $$\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$$ and $$\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$$.

 

[Prove It]

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.

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.

$\displaystyle \begin{align*} \frac{\mathrm{e}^{\mathrm{i}\,\sqrt{x}}-1}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} &= \frac{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1 - 2}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} \\ &= 1 - \frac{2}{\mathrm{e}^{\mathrm{i}\,\sqrt{x}} + 1} \\ &= 1 - \frac{2}{1 + \cos{ \left( \sqrt{x} \right) } + \mathrm{i} \sin{\left( \sqrt{x} \right) }} \\ &= 1 - \frac{2\,\left[ 1 + \cos{ \left( \sqrt{x} \right) } - \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] }{\left[ 1 + \cos{\left( \sqrt{x} \right) } + \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] \left[ 1 + \cos{ \left( \sqrt{x} \right) } - \mathrm{i}\sin{ \left( \sqrt{x} \right) } \right] } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right)
} - 2\,\mathrm{i} \sin{ \left( \sqrt{x} \right) } }{ \left[ 1 + \cos{ \left( \sqrt{x} \right) } \right] ^2 + \sin^2{ \left( \sqrt{x} \right) } } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right) } - 2\,\mathrm{i}\sin{\left( \sqrt{x} \right) }}{1 + 2\cos{ \left( \sqrt{x} \right) } + \cos^2{ \left( \sqrt{x} \right) } + \sin^2{ \left( \sqrt{x} \right) } } \\ &= 1 - \frac{2 + 2\cos{\left( \sqrt{x} \right) } - 2\,\mathrm{i}\sin{\left( \sqrt{x} \right) }}{2 + 2\cos{\left( \sqrt{x} \right) }} \\ &= 1 - \left[ 1 - \frac{2\,\mathrm{i}\sin{\left( \sqrt{x} \right) } }{2\,\left[ 1 + \cos{\left( \sqrt{x} \right) } \right] } \right] \\ &= \mathrm{i}\,\left[ \frac{\sin{\left( \sqrt{x} \right) }}{1 + \cos{ \left( \sqrt{x} \right) } } \right] \\ &= \mathrm{i}\,\left[ \frac{2\sin{\left( \frac{\sqrt{x}}{2} \right) } \cos{\left( \frac{\sqrt{x}}{2} \right) }}{1 + 2\cos^2{\left( \frac{\sqrt{x}}{2} \right) } - 1 } \right] \\ &= \mathrm{i}\,\left[ \frac{2\sin{\left( \frac{\sqrt{x}}{2} \right) }\cos{\left( \frac{\sqrt{x}}{2}\right) }}{2\cos^2{\left( \frac{\sqrt{x}}{2} \right) }} \right] \\ &= \mathrm{i}\tan{ \left( \frac{\sqrt{x}}{2} \right) } \end{align*}$

 

[Greg]

$$\dfrac{e^{i\sqrt x}-1}{e^{i\sqrt x}+1}=\dfrac{e^{i\sqrt x/2}-e^{-i\sqrt x/2}}{e^{i\sqrt x/2}+e^{-i\sqrt x/2}}=i\dfrac{\sin\dfrac{\sqrt x}{2}}{\cos\dfrac{\sqrt x}{2}}=i\tan\dfrac{\sqrt x}{2}$$

 

[Guest2]

Thanks, guys. I appreciate this. :D

$$\dfrac{e^{i\sqrt x}-1}{e^{i\sqrt x}+1}=\dfrac{e^{i\sqrt x/2}-e^{-i\sqrt x/2}}{e^{i\sqrt x/2}+e^{-i\sqrt x/2}}=i\dfrac{\sin\dfrac{\sqrt x}{2}}{\cos\dfrac{\sqrt x}{2}}=i\tan\dfrac{\sqrt x}{2}$$

I must remember this one for my exam. Sweet and short!