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Question Number 139501 by aliibrahim1 last updated on 28/Apr/21

Answered by mr W last updated on 28/Apr/21

$${(x+3i)}^{100}=-1=e^{(2k+1)\pi i}$$$$x_k+3i=e^{\frac{(2k+1)\pi }{100}i}$$$$x_k=e^{\frac{(2k+1)\pi }{100}i}-3i=\cos \frac{(2k+1)\pi }{100}+(\sin \frac{(2k+1)\pi }{100}-3)i$$$$=\sqrt{10-6\sin \frac{(2k+1)\pi }{100}}{e}^{-{\tan }^{-1}(\frac{3}{\cos \frac{(2k+1)\pi }{100}}-\tan \frac{(2k+1)\pi }{100})i}$$$$\mathrm{\Pi }x=\underset{k=0}{\prod ^{99}}\sqrt{10-6\sin \frac{(2k+1)\pi }{100}}{e}^{-{\tan }^{-1}(\frac{3}{\cos \frac{(2k+1)\pi }{100}}-\tan \frac{(2k+1)\pi }{100})i}$$$$=\left(\underset{k=0}{\prod ^{99}}\sqrt{10-6\sin \frac{(2k+1)\pi }{100}}\right)e^{-\underset{k=0}{\sum ^{99}}{\tan }^{-1}(\frac{3}{\cos \frac{(2k+1)\pi }{100}}-\tan \frac{(2k+1)\pi }{100})i}$$$$......$$

Commented by aliibrahim1 last updated on 28/Apr/21

$$wooowthxsir$$

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