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 #1
I started of with attempting to convert the numerator first
$  1 + i  = \sqrt{1^2+i^2}$
$= \sqrt{11} = 0$ ? this is wrong obviously, i dont see why its $\sqrt{2}$
for the second part
$ \sqrt{3}  i= \sqrt{3+1} = 2$
$ x = r \cos\theta$ $ y = r\sin\theta$
$x = 2\cos\theta$ $ y=2\sin\theta$
then $\theta = \frac{\pi}{3} and \frac{\pi}{6}$
$ = 2(\cos\frac{\pi}{3} + \sin\frac{\pi}{6} = 2cis(\frac{\pi}{3})$ I don't see why this is wrong either
$  1 + i  = \sqrt{1^2+i^2}$
$= \sqrt{11} = 0$ ? this is wrong obviously, i dont see why its $\sqrt{2}$
for the second part
$ \sqrt{3}  i= \sqrt{3+1} = 2$
$ x = r \cos\theta$ $ y = r\sin\theta$
$x = 2\cos\theta$ $ y=2\sin\theta$
then $\theta = \frac{\pi}{3} and \frac{\pi}{6}$
$ = 2(\cos\frac{\pi}{3} + \sin\frac{\pi}{6} = 2cis(\frac{\pi}{3})$ I don't see why this is wrong either
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