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##### Quiz 3 / Quiz3 problem 6E

« Last post by**Xuefen luo**on

*October 18, 2020, 03:52:58 PM*»

Problem: Compute the following line integral:

\begin{align*}\int_{\gamma}^{} |z|^2 \,dz, \end{align*}

where $\gamma$ is the line segment from 2 to 3 + i

\begin{align*}

\end{align*}

Answer:

\begin{align*}

\gamma(t)&= (1-t)z_0+tz_1\\

&= 2(1-t)+(3+i)t\\

&= 2-2t+3t+it\\

&= 2+t+it \ \ \ \ (0 \leq t \leq 1)\\

\\

\gamma'(t)&=1+i\\

\\

Let\ f(z)=|z|^2\\

\\

f(\gamma(t))&= |\gamma(t)|^2\\

&=(\sqrt{(2+t)^2+t^2})^2\\

&=4+4t+t^2+t^2\\

&=2t^2+4t+4\\\

\\

\int_{\gamma}^{} |z|^2 \,dz &= \int_{\gamma}^{} f(z) \,dz\\

&=\int_{0}^{1} f(\gamma(t))\ \gamma'(t) \,dt\\

&=\int_{0}^{1} (2t^2+4t+4) (1+i) \,dt\\

&=(1+i)\int_{0}^{1} (2t^2+4t+4) \,dt \\

&=(1+i)(\frac{2}{3}t^3 + 2t^2 +4t)|_{0}^{1}\\

&=(1+i)(\frac{2}{3} + 2 +4)\\

&=\frac{20}{3}(1+i)

\end{align*}

\begin{align*}\int_{\gamma}^{} |z|^2 \,dz, \end{align*}

where $\gamma$ is the line segment from 2 to 3 + i

\begin{align*}

\end{align*}

Answer:

\begin{align*}

\gamma(t)&= (1-t)z_0+tz_1\\

&= 2(1-t)+(3+i)t\\

&= 2-2t+3t+it\\

&= 2+t+it \ \ \ \ (0 \leq t \leq 1)\\

\\

\gamma'(t)&=1+i\\

\\

Let\ f(z)=|z|^2\\

\\

f(\gamma(t))&= |\gamma(t)|^2\\

&=(\sqrt{(2+t)^2+t^2})^2\\

&=4+4t+t^2+t^2\\

&=2t^2+4t+4\\\

\\

\int_{\gamma}^{} |z|^2 \,dz &= \int_{\gamma}^{} f(z) \,dz\\

&=\int_{0}^{1} f(\gamma(t))\ \gamma'(t) \,dt\\

&=\int_{0}^{1} (2t^2+4t+4) (1+i) \,dt\\

&=(1+i)\int_{0}^{1} (2t^2+4t+4) \,dt \\

&=(1+i)(\frac{2}{3}t^3 + 2t^2 +4t)|_{0}^{1}\\

&=(1+i)(\frac{2}{3} + 2 +4)\\

&=\frac{20}{3}(1+i)

\end{align*}