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Question

The product of all values of cosα+ιsinα35


A

1

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B

cos(α)+isin(α)

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C

cos(3α)+isin(3α)

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D

cos(5α)+isin(α)

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Solution

The correct option is C

cos(3α)+isin(3α)


Explanation for the correct option:

Given, cosα+ιsinα35
But, cosα+isinα35=cosα+isinα315

From de Moivre's theorem,
cosα+isinαn=cosnα+isinnα where n is an integer

Thus,
cosα+isinα315=cos3α+isin3α15

Also by de Moivre's theorem,
cosα+isinα1n=cos2kπ+αn+isin2kπ+αn where n is an integer and k=0,1,2,,n-1

Thus, cos3α+isin3α15=cos2kπ+3α5+isin2kπ+3α5,k=0,1,2,3,4
Thus, the values of cosα+ιsinα35 are cos3α5+isin3α5, cos2π+3α5+isin2π+3α5,cos4π+3α5+isin4π+3α5 cos6π+3α5+isin6π+3α5 and cos8π+3α5+isin8π+3α5
Their product is,

=cos3α5+isin3α5×cos2π+3α5+isin2π+3α5×cos4π+3α5+isin4π+3α5×cos6π+3α5+isin6π+3α5×cos8π+3α5+isin8π+3α5=cos3α5+2π5+3α5+4π5+3α5+6π5+3α5+8π5+3α5+isin3α5+2π5+3α5+4π5+3α5+6π5+3α5+8π5+3α5=cos20π5+5×3α5+isin20π5+5×3α5=cos(4π+3α)+isin(4π+3α)=cos(3α)+isin(3α)

Hence option (C) is correct.


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