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Could someone help me through this problem? 

Consider the integral $$\int_{0}^{b} \frac{1}{z^{\frac{1}{2}}}\, dz, b>0 $$ Let $z^{\frac{1}{2}}$ have a branch cut along the positive real axis. Show that the value of the integral obtained by integrating along the top half of the cut is exactly minus that obtained by integrating along the bottom half of the cut. What is the difference between taking the principal versus the second branch of $z^{\frac{1}{2}}$?

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In this problem we look at the upper bound of the integration $b$ and how it changes depending on where we integrate along the branch cut. If we integrate along the top of the branch cut then we know $$ |b|=b \quad and \quad arg(b)=0.$$ Therefore, $$\int_{0}^{b} \frac{1}{z^{\frac{1}{2}}}\, dz =2b^{\frac{1}{2}}.$$ However, if we integrate along the bottom of the branch cut we have $$ |b|=b \quad and \quad arg(b)=2\pi.$$ Therefore, $$\int_{0}^{b} \frac{1}{z^{\frac{1}{2}}}\, dz =2\left(be^{i2\pi}\right)^{\frac{1}{2}}=2b^{\frac{1}{2}}e^{i\pi} = -2b^{\frac{1}{2}}.$$ So the value of the integral is the same but the sign is opposite. With this in mind we can answer the part about integrating over the second branch as along the top of the second branch of $z^{\frac{1}{2}}$ we have $$|b|=b\quad and \quad arg(b)=2\pi,$$ and along the bottom of the second branch we have $$|b|=b\quad and \quad arg(b)=4\pi.$$ We then perform the same integration as before and notice that now the result is negative along the top of the second branch and positive along the bottom of the second branch.

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