Essentially the title.
This is a homework problem, and I tried to do it by considering it as a $1 ,\sqrt{3},2$ triangle as the angle is $\frac{\pi}{3}$, and thus $b$ would be $\sqrt{3}$. However, the answer section in the back of the textbook says the answer is $\frac{1}{\sqrt{3}}$, and I am confused why this is the case.
Thanks
From the polar form of a complex number we know that squaring will square the radius and double the angle. This gives us $\arg (1+bi) =\frac{\pi}{6}$ and since we can calculate $|1+bi| = \sqrt{1 + b^2}$ we now have the argument and the radius.
Then by converting back to cartesian coordinates we find that $x=1=\cos(\frac{\pi}{6})\sqrt{1 + b^2}$ and it follows that $$b = \sqrt{\cos(\frac{\pi}{6})^{-2}-1}$$ from here simply substitute $\cos{\frac{\pi}{6}}= \frac{\sqrt{3}}{2}$ and we finally have that $b =\frac{1}{\sqrt{3}}$ as the book suggests.