I have been given the following formula: $$ f(z) = \sqrt{(z^2+1)} = \sqrt{(z+i)(z-i)} $$
And I have to prove that i and -i are two branch points: if you make a circle around either of those points in the domain you do not end up in the same spot as where you started.
So I try to do that and the first step for me would be to find a z for which this goes. The easy one seems to be $$ z = e^{xi} + i $$ But this does not work: $f(e^{0i} + i) = f(e^{2 \pi i} + i) = \sqrt{(1+2i)} = \sqrt{(1+2i)(1)}$
Is my Z function wrong, am I missing something? Which Z function will show me that $z = z_1$ does not lead to $f(z) = f(z_1)$ when going around i and -i?
Funny thing: the answer is in my question, but I made a mistake while calculating. First, $f(z)$: $$ f(z) = \sqrt{(z^2+1)} = \sqrt{(z+i)(z-i)} = \sqrt{z+i}\sqrt{z-i} $$ Then: $e^{xi} + i$, so lets compare $z_1 = e^{0i} + i$ and $z_2 = e^{(0 + 2\pi)i} + i = e^{0i}e^{2\pi i} + i$ $$ f(e^{0i} + i) = (\sqrt{e^{0i} + 2i})e^{\frac{0}{2}i} = \sqrt{1 + 2i} $$ $$ f(e^{0i}e^{2\pi i} + i) = (\sqrt{e^{0i}e^{2\pi i} + 2i})e^{\frac{0}{2}i}e^{\pi i} = -\sqrt{1 + 2i} $$
Hence, branch point at $z = i$. Same thing goes for $z = -i$