Solve and write in standard form: z = (-√3 + i )^5

Hello from MrBillDoesMath!

16 ( sqrt(3) + i)

Discussion:

As  (-sqrt(3) + i) in polar form is  2 ( cos(150) + isin(150) )

        (Angles shown in degrees for simplicity.

z = ( -sqrt(3) + i)          

z = 2 ( cos(150) + i sin(150) )         =>

z^5 = 2^5 ( cosi(150) + i sin(150))  ^5     (*)

De Moivre's theorem for complex number gives

(cos x+isin x)^n =  cos(nx)+i sin(nx)

Evaluating the rhs of (*) gives

z^5 =

      = 2^5 ( cos(150*5) + i sin (150*5)

      = 32 (   cos(750)     + i sin(750) )

      = 32 (  sqrt(3) /2     + i (1/2) )

      = 16 sqrt(3) + 16 i

     = 16 ( sqrt(3) + i)

Thank you,

MrB

A and E.

Step-by-step explanation:

For a function of the form:

F(x) = h(x)/g(x)

It only has a vertical asymptote in the points where the denominator is equal to zero, this means that we must have g(x) = 0.

In this case, we have:

F(x) = (x-4)(x + 2)

As this function has no denominator, this will not have any vertical asymptote, then i suppose that the actual function is:

Using the same notation as above, we have:

g(x) = (x - 4)*(x + 2)

Then the vertical asympotes of F(x) will be on the values of x such that:

g(x) = 0.

As g(x) is already written in factorized form, we know that the zeros will be at the values of x that make one of the terms inside the parentheses equal to zero, these values are:

x = 4:

g(4) = (4 - 4)*(4 + 2) = (0)*6 = 0

and at x = -2

g(-2) = (-2 - 4)*(-2 + 2) = -6*0 = 0

Then the two asymptotes of F(x) are at x = 4, and x = -2

The correct options are:

A and E.