Modulus of integral

Calculus Level 3

z = π 2 π e i θ cos θ d θ \large z = \int_\frac \pi 2^\pi e^{i \theta} \cos \theta \ d\theta

For complex number z z as defined above, find z |z| . Give your answer to 3 decimal places.

Notations:


The answer is 0.931.

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1 solution

Chew-Seong Cheong
Sep 20, 2018

Relevant wiki: Euler's Formula

z = π 2 π cos θ e i θ d θ By Euler’s formula e i x = cos x + i sin x = π 2 π cos θ ( cos θ + i sin θ ) d θ = π 2 π ( cos 2 θ + i sin θ cos θ ) d θ = π 2 π 1 2 ( 1 + cos 2 θ + i sin 2 θ ) d θ = 1 2 [ θ + sin 2 θ 2 i cos 2 θ 2 ] π 2 π = π 4 i 2 \begin{aligned} z & = \int_\frac \pi 2^\pi \cos \theta {\color{#3D99F6}e^{i\theta}} d\theta & \small \color{#3D99F6} \text{By Euler's formula }e^{ix} = \cos x + i \sin x \\ & = \int_\frac \pi 2^\pi \cos \theta {\color{#3D99F6}(\cos \theta + i \sin \theta)} d\theta \\ & = \int_\frac \pi 2^\pi \left(\cos^2 \theta + i \sin \theta \cos \theta \right) d\theta \\ & = \int_\frac \pi 2^\pi \frac 12 \left(1 + \cos 2 \theta + i \sin 2 \theta \right) d\theta \\ & = \frac 12 \left[ \theta + \frac {\sin 2\theta}2 - i \frac {\cos 2\theta}2\right]_\frac \pi 2^\pi \\ & = \frac \pi 4 - \frac i2 \end{aligned}

z = π 4 i 2 = π 2 16 + 1 4 0.931 \begin{aligned} \implies |z| & = \left|\frac \pi 4 - \frac i2\right| = \sqrt{\frac {\pi^2}{16}+\frac 14} \approx \boxed{0.931} \end{aligned}

@Charley Feng , it may be better to use the standard z z to denote a complex number instead of c c which usually denotes a constant. Also just mention "Find z |z| " instead of "What is the distance from the origin to c c ?"

Chew-Seong Cheong - 2 years, 8 months ago

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Ok thanks for the advice, I changed it to z. I just wanted to be original with the letters haha. And yea, I wanted to word it like that to make you think a little bit rather than just stating it plainly as the modulus of z.

Charley Shi - 2 years, 8 months ago

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I have edited the problem wording for you.

Chew-Seong Cheong - 2 years, 8 months ago

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@Chew-Seong Cheong ok thank you.

Charley Shi - 2 years, 8 months ago

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