Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 $\frac{1}{2}$.

share|improve this answer
add comment

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 $t=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.