# Upper Bound of Line Integral

Obtain an Upper Bound for |$\int_\gamma (z^2 + 2)^{-1}$| when $\gamma$ is the line segment from 0 to 1 + i.

So far I have determined $\gamma = te^{i\pi/4}$ and the length of $\gamma = \sqrt{2}$

Now using the estimation lemma I have to determine a value for M but I am not sure how to go about this.

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Let $z=t+it$, $\ \$ $0\le t \le 1$. Then $$|\frac{1}{1+z^2}|=|\frac{1}{2+t^2+2it-t^2}|=|\frac{1}{2+2it}|=\frac{1}{ 2 \sqrt{1+t^2}} \ .$$ The maximum value is at $t=0$.
You want the choose the largest value of the function $$\frac{1}{|z^2 + 2|}$$ on the line from $0$ to $1+i$. One useful way to view these problems is to geometrically interpret them.
The image of the line $\gamma$ under the function $z^2$ is given by the vertical line from $0$ to $2$. How do I know this? The line $\gamma$ forms angle $\frac{\pi}{4}$ with the real axis. The function $z^2$ doubles the angle as measured from the real axis so that $\gamma$ is taken to a line with angle $\frac{\pi}{2}$, i.e. the vertical line. The function also squares magnitudes so that the endpoint with original magnitude $\sqrt{2}$ now has magnitude $\sqrt{2}^2 = 2$. It is now easy to see that the value of $|z^2 + 2|$ will be minimized when $z=0$ so that the maximum of $$\frac{1}{|z^2 + 2|}$$ on $\gamma$ is precisely $\frac{1}{2}$.