GATE Overflow Test Series | Mock GATE | Test 1 | Question: 12

Question

Assuming $i= \sqrt{-1}$ and $t$ is a real number , $$I =\int_{0 }^{\frac{\pi}{3}}e^{it}dt$$

• A. $\frac{\sqrt{3}}{2} + i\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2} - i\frac{1}{2}$
• C. $\frac{1}{2} + i\frac{\sqrt{3}}{2}$
• D. $\frac{1}{2} + \left(1- \frac{\sqrt{3}}{2}\right)$

Answer 1

$I = \displaystyle{}\int_{0 }^{\frac{\pi}{3}}e^{it}dt = \left[\frac{e^{it}}{i}\right]_{0}^{\frac{\pi}{3}}$

Put, $e^{it} = \cos t + i\sin t$

$ = \left[\dfrac{\cos t + i\sin t}{i}\right]_{0}^{\frac{\pi}{3}}$

$ = \frac{1}{i}\left[\frac{1}{2} + i\frac{\sqrt{3}}{2}  - 1\right]$

$ = \frac{1}{i}\left[-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right] = -i\left[-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right]=  \frac{\sqrt{3}}{2} + i\frac{1}{2}.\quad [\because i^{2}=-1]$

Comments

Lakshman Patel RJIT

can’t understand how 1/i becomes -i. Can u please explain

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We know that, $i = \sqrt{-1} \implies i^{2} = -1$

Now, $\dfrac{1}{i} = \dfrac{1}{i} \times \dfrac{i}{i} = \dfrac{i}{i^{2}} = \dfrac{i}{-1} = -i.$

See for more info https://brilliant.org/wiki/complex-numbers/

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why  we should take e power i*t as cos t + i sin t  any reason for specific equation?

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marked b in a hurry 😭