4) Evaluate - dz, where y is the line from 1 → 1+ iSy'Y z

Answered: 4) Evaluate , -dz, where y is the line…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

4) Evaluate - dz, where y is the line from 1 → 1+ i
Sy
'Y z

Expert Solution

Step 1

In complex integration, we have to convert our given question into complex terms. But in our question lines is from 1 to 1+i and from these, we can say these curve does not contain the origin, so we can use antiderivative here without transforming into the complex form.

The line is from 1 to 1+i, that is (1,0) to (1,1) in a complex plane. So it does not have (0,0), that's why I said above it does not contain origin so we can use antiderivative directly here.

Step 2

We have $\int_{1}^{1 + i} 1 / z d z$

$\int_{1}^{1 + i} 1 / z d z =$ ${[\begin{matrix} \log & z \end{matrix}]}_{1}^{1 + i}$ = log (1+i) - log(1) = log(1+i)

We know log(1+i) = log $\sqrt{2}$ + $i * π / 4$ .

Explanation for log(1+i):

x+iy = re^{i $θ$}

x + iy = 1+i, so in our case x=1 and y =i.

r= $\sqrt{(x^{2} + y^{2})} = \sqrt{(1^{2} + 1^{2})} = \sqrt{2}$

$θ = \tan^{- 1} y$

In our case y =1, so $θ = \tan^{- 1} 1 = π / 4$

1+i = $\sqrt{2} e^{i π / 4}$

log(1+i) = log( $\sqrt{2} e^{i π / 4}$ ) = log $\sqrt{2} + \log e^{i π / 4}$ = log $\sqrt{2} + i π / 4$ .

The real part is log $\sqrt{2}$ and the imaginary part is $π / 4$ .

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