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Evaluate $ \int_c x dz $ from $-4$ to $4$ along countour

countour polygon

It will take me a few hours to draw(type ) on latex


got the answer that its 0 but it had diff solution of back of book

First find parameterizations for diagram. The general equation for line of two complex points is $$z(t)=z_1+(z_2-z_1)t \hspace{ .5cm} t\in[0,1] $$ For $C_1$ $z_1=-4,x_2=-4+4i $ $t\in[0,1]$ $$\begin{aligned} z_1(t)&= z_1+(z_2-z_1)t \\ &=-4+(4i)t \\ &=-4+4it \end{aligned}$$

For $C_2$ $z_1=-4+4i$,$z_2=4+4i$ $$\begin{aligned} z_1(t)&= z_1+(z_2-z_1)t \\ &=(-4+4i)+((4+4i)-(-4+4i))t \\ &=-4+4i+8t \\ &=(-4+8t)+4i \end{aligned}$$

For $C_3$ $z_1=4+4i$ and $z_2=4$ $$\begin{aligned} z_1(t)&= z_1+(z_2-z_1)t \\ &=(4+4i)+((4)-(4+4i))t \\ &=4+i(4+4t) \\ &=4+i4(1+t) \\ &=4+4i+4it \end{aligned}$$

Trying to integrate $$ \int_c x dz = \int_{c_{1}} x dz+ \int_{c_{2}} x dz+\int_{c_{3}} x dz$$ Using $\int_c f(z)dz=\int^a_b f(z(t))z'(t)dt $ $$\begin{aligned} \int_{c_{1}} x dz &= \int^1_0 (-4 *i)dt \\ \int_{c_{2}} x dz &= \int^1_0 (4+8t)8dt \\ \int_{c_{3}} x dz &= \int^1_0 (4)(4i)dt \end{aligned} $$ so $$ \begin{aligned} \int_c x dz &= \int_{c_{1}} x dz+ \int_{c_{2}} x dz+\int_{c_{3}} x dz = \int_{c_{2}} x dz =\int^1_0 (4+8t)8dt =8[-4t+8/2 t^2]^1_0 =0 \end{aligned}$$

so $\int_c f(z)dz=0$ but the answer back of book is $-32i$ something is off If i had to guess what I am doing wrong is not seeing what is $f(z(t))$ which i think should be x(t) if $z(t)=x(t)+iy(t)$

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Your integral over $C_1$ is missing $x(t) = 4$ and thus should be $\int_{0}^1 4 \cdot (-4i) \text{d}t$.
Your last integral is in the wrong direction, as $z_1 = 4+4i - 4it$.
So instead of $\int_{C_1} x \text{d}z = - \int_{C_3} x \text{d}z$, you need to integrate both integrals, and add the outcomes to the result, which gives $-32 i$.

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