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I am new to quantum computing. Can someone help me in calculating state of the qubit when $\alpha = \frac{3}{5}e^{i\pi/7}$ and $\beta = -\frac{4i}{5}$.

Thanks

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    $\begingroup$ Assuming the qubit is given by $\alpha\vert 0\rangle + \beta\vert 1\rangle$, this question is trivial right? However, from your question it does not become clear what you are really asking. There seems to be little calculation involved here. $\endgroup$ – nippon May 11 '20 at 7:48
  • $\begingroup$ @nippon are you able to see deleted posts? A deleted answer by Mariia has a comment by the original author of the question, that clarifies a little bit, what they want. $\endgroup$ – user1271772 May 11 '20 at 14:07
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The state is usually just expressed as $$ \alpha|0\rangle+\beta|1\rangle, $$ which is straightforward given that you're stating $\alpha$ and $\beta$.

However, if what you actually want is a geoemtric interpretation, that means mapping your system to something of the form $$ e^{i\gamma}(\cos\frac{\theta}{2}|0\rangle+\sin\frac{\theta}{2}e^{i\phi}|1\rangle) $$ where you want the values of $\theta$ and $\phi$. In this case, you immediately have $\gamma=\pi/7$ and $\cos\frac{\theta}{2}=\frac{3}{5}$. You want to be a little careful in your choice of $\sin\frac{\theta}{2}$ - you probably want it to be positive (as was also true of cos), so $\sin\frac{\theta}{2}=\frac{4}{5}$. Hence, $$ \sin\theta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}. $$ You can numerically evaluate this of you want, but it's probably better to leave it just like that.

Now we need to take care of $\phi$. We have $$ e^{i\phi}=-ie^{-i\pi/7}=e^{-i\pi/7-i\pi/2}, $$ so we conclude that $$ \phi=2\pi-\frac{\pi}{7}-\frac{\pi}{2}=\frac{19}{14}\pi, $$ since you're probably expected to give a value of $\phi$ in the range 0 to $2\pi$.

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