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If theta=pi/6 then the 10^(th) term of t...

If `theta=pi/6` then the `10^(th)` term of the series : `1+(costheta+isintheta)^1+(costheta+isintheta)^2+........`

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If theta=pi//6 then the 10th term of the series 1+(costheta+isintheta)+(costheta+isintheta)^(2)+...... is

(cos3theta+isin3theta)^5/(costheta+isintheta)^6

If theta=pi/6, then the 10th term of the series 1+(costheta+isintheta)+(cos theta+i sin theta)^2+ …. Is

(1-costheta+isintheta)^6 =

(1-costheta+isintheta)^(6)

(1-costheta+isintheta)^8 =

(1+costheta-isintheta)^(n)

(1+costheta-isintheta)^4 =

(costheta+isintheta)^(10)+(costheta-isintheta)^(10)

|1+costheta+isintheta|=