If `z and bar z` represent adjacent vertices of a regular polygon of `n` sides where centre is origin and if `(Im(z))/(Re(z)) = sqrt(2) - 1`, then `n` is equal to:

Options

A. 8
B. 16
C. 24
D. 32

Answer

A

Explanation

Clearly, `barz=ze^(i2pi//n`. Let `z=r,e^(itheta)`. Then, `barz=re^(-theta)`.
`rArr re^(-itheta)e^(i2pi//n) rArr e^(-i2theta)=e^(i2pi/n) rArr theta=-pi/n`

Now, `|("Im"(z))/("Re"(z))|=sqrt(2)-1`
`rArr |(rsintheta)/(rcostheta)|=sqrt(2)-1`
`rArr tanpi/2 =sqrt(2)-1 rArr tan pi/n = tanpi/8 rArr n=8`