Another Another Complex numbers question (1 Viewer)

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khorne

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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

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(k = 0,1,2,3,4)
min value of occurs when k = 1



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