Answer:

1) Since w = cos 2π/3 + i sin 2π/3 , this means it's one of the cube root of unity and then we know the identity

1+w+w^2 = 0

Now, we'll modify the determinant a little bit, First transformation, adding C2 and C3 to C1

z+1+w+w^2 w. w^2

z+1+w+w^2 z+w^2 1

z+1+w+w^2. 1. z+w

=>. z. w. w^2.

z. z+w^2. 1

z. 1. z+w

Taking out z common from first column, and then solving the determinant along C1 we get,

z { (z+w^2)(z+w)-1) - z(z+w) + w^2 + w -w^2(z+w^2) }

= 0

=> z ( z^2 + wz + w^2 z + w^3 - z^2 - wz + w^2 + w - w^2z - w^4 ) = 0

=> z ( w^3 + w + w^2 - w) = 0

or, z(-w) = 0

or, z = 0

So, there's only one real root of the determinant equation that is z=0

2) By DeMoivre' s formula

(cis π/3)^3/4 = z

or,

z^4 = (cis π/3)^3 = cis π+2nπ = -1 +0i = -1

So,

z = (cis π+2nπ)^1/4 = cis (π/4+nπ/2)

=> z1 = cis π/4 , z2 = cis 3π/4 , z3 = cis 5π/4 , z4 = cis 7π/4

So, z1*z2*z3*z4

= e^iπ/4 * e^i3π/4 * e^i5π/4 * e^i7π/4

= e^i(4π) = 1

So, product of all these roots will be 1

Hope this helps you !

Question-1

We know,

Here,

We have add the transformations of determinant.

z is common in C₁ .

So,

solving the determinant along C₁ ,

Therefore z = 0 is the only root.

• Correct Option - Option 1

_________________________

Question-2

Here,

So,

Therefore,

• Correct Option - Option 3