Another Another Complex numbers question (1 Viewer)

Hermes1 · Jul 23, 2011

$\frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta } = \sin \theta + i \cos \theta$

Hence find the smallest value of $\theta$ such that:

$(1+\sin \theta + i\cos \theta )^{5} + i(1 + \sin \theta - i\cos \theta )^{5} = 0$

khorne · Jul 23, 2011

divide both sides by (1+sinx -icosx), then you get (sinx + icosx)^5 + i = 0

-i = cis(-pi/2)

So cis5x = cis(-pi/2)
5x = -pi/2 + 2k(pi)
x = (-pi+4k(pi))/10, k = 0,1,2,3... etc

And the smallest is at k = 1 (positive smallest I assume), so 3pi/10

Hermes1 · Jul 23, 2011

$(\sin \theta +i\cos \theta )^{5}=-i$
$(\cos (\frac{\pi }{2}-\theta )+i\sin (\frac{\pi }{2}-\theta ) )^{5}=-i$
$\therefore cis 5(\frac{\pi }{2}-\theta )= cis -\frac{\pi }{2}+2k\pi$ (k = 0,1,2,3,4)
min value of $\theta$ occurs when k = 1
$\therefore \frac{\pi }{2}-\theta = \frac{3\pi }{10}$
$\therefore \theta = \frac{\pi }{5}$

khorne dont you have to change the sin x + i cos x into cis form before equating with cis -pi/2

powlmao · Jul 23, 2011

arj1 said:
$(\sin \theta +i\cos \theta )^{5}=-i$
$(\cos (\frac{\pi }{2}-\theta )+i\sin (\frac{\pi }{2}-\theta ) )^{5}=-i$
$\therefore cis 5(\frac{\pi }{2}-\theta )= cis -\frac{\pi }{2}+2k\pi$ (k = 0,1,2,3,4)
min value of $\theta$ occurs when k = 1
$\therefore \frac{\pi }{2}-\theta = \frac{3\pi }{10}$
$\therefore \theta = \frac{\pi }{5}$

Negetive Imaginary number !!!

khorne · Jul 23, 2011

arj1 said:
$(\sin \theta +i\cos \theta )^{5}=-i$
$(\cos (\frac{\pi }{2}-\theta )+i\sin (\frac{\pi }{2}-\theta ) )^{5}=-i$
$\therefore cis 5(\frac{\pi }{2}-\theta )= cis -\frac{\pi }{2}+2k\pi$ (k = 0,1,2,3,4)
min value of $\theta$ occurs when k = 1
$\therefore \frac{\pi }{2}-\theta = \frac{3\pi }{10}$
$\therefore \theta = \frac{\pi }{5}$

khorne dont you have to change the sin x + i cos x into cis form before equating with cis -pi/2

yes, missed that

Hermes1 · Jul 23, 2011

khorne said:
yes, missed that

tnx for ur help

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