locus of the complex number

Its a circle. I know you will curse me for not providing the solution. But, you could try to prove this and also I want the students to have a go at it.

thanks sir , i wanted to verify as answer is not given
sir is it a circle of radius 3 and centre shifted by two units in + x axis
?

shall i give the solution??

wont jus multiplying and dividing with conjugate of teh complex no 2+cosθ+isinθ help....

after that put z=x+iy and then equate real and imaginary parts.....and then eliminate cosθ from both eq to get an eq in x and y

z = \frac{3}{2+\cos \theta + i \sin \theta} \Rightarrow \frac{3}{z} -2 = \cos \theta + i \sin \theta

\Rightarrow \left|\frac{3}{z} -2 \right|= 1 \Rightarrow |z| = 2 \left|z - \frac{3}{2} \right|

Thus, we are looking for points whose distances from (0,0) and (3/2,0) are in a fixed ratio which is nothing but a circle

(That's all they need, radius centre etc. are extraneous info)

for 2nd q u got to prove this

cos(n\theta) =2^{n-1}\left(cos\theta -cos\frac{\pi }{2n} \right)\left(cos\theta -cos\frac{3\pi }{2n} \right)..........\left(cos\theta -cos\frac{(2n-1)\pi }{2n} \right)

for this q

put θ=pie

n=5

ans 12⁴