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If e^{i\alpha}= \cos \alpha+i \sin \alpha , then for the \Delta ABC, e^{iA} \times e^{iB} \times e^{iC}   is:

  • Option 1)

    -i

  • Option 2)

    1

  • Option 3)

    -1

  • Option 4)

    None of these.

 

Answers (1)

As we learnt in 

Euler's Form of a Complex number -

z=re^{i\theta}

- wherein

r denotes modulus of z and \theta denotes argument of z.

 

 \\e^{iA}\times e^{iB}\times e^{iC}\\*\\*=e^{i\left ( A+B+C \right )}=e^{i \pi}=cos \pi+isin \pi=-1

Since sum of angles=\pi


Option 1)

-i

Incorrect

Option 2)

1

Incorrect

Option 3)

-1

Correct

Option 4)

None of these.

Incorrect

Posted by

Vakul

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