7 marks) Let α = −1 + j and β = 4 − j. In each case, solve for z.(a) z − β∗ = Im( α22β + j)(b) z4 = α

(a) z - β* = Im( α^2/2β + j)

First, let's find the complex conjugate of β, denoted as β*. Since β = 4 - j, β* = 4 + j.

Next, let's calculate α^2/2β. Since α = -1 + j and β = 4 - j, we have:

α^2 = (-1 + j)^2 = 1 - 2j + j^2 = 1 - 2j - 1 = -2j.

So, α^2/2β = -2j/(2*(4 - j)) = -j/(4 - j).

Now, let's calculate the imaginary part of α^2/2β + j. Since α^2/2β = -j/(4 - j), we have:

Im(α^2/2β + j) = Im(-j/(4 - j) + j) = Im((-j + j*(4 - j))/(4 - j)) = Im((j^2*(4 - j))/(4 - j)) = Im((j^2*4 - j^3)/(4 - j)).

Since j^2 = -1 and j^3 = -j, we have:

Im((j^2*4 - j^3)/(4 - j)) = Im((-4 - j)/(4 - j)) = -4.

Finally, let's solve for z. Since z - β* = Im( α^2/2β + j), we have:

z = β* + Im( α^2/2β + j) = (4 + j) + (-4) = j.

So, the solution to part (a) is z = j.

(b) z^4 = α

To solve this equation, we need to find the fourth roots of α = -1 + j.

The fourth roots of a complex number can be found using the formula z = r^(1/n)(cos(θ/n) + jsin(θ/n)), where r is the magnitude of the complex number, θ is the argument of the complex number, and n is the root we want to find.

First, let's find the magnitude of α. Since α = -1 + j, we have:

r = |α| = sqrt((-1)^2 + 1^2) = sqrt(2).

Next, let's find the argument of α. Since α = -1 + j, we have:

θ = atan2(1, -1) = 3π/4.

Now, let's find the fourth roots of α. Since n = 4, we have:

z = sqrt(2)^(1/4)(cos(3π/16) + jsin(3π/16)), z = sqrt(2)^(1/4)(cos(11π/16) + jsin(11π/16)), z = sqrt(2)^(1/4)(cos(19π/16) + jsin(19π/16)), z = sqrt(2)^(1/4)(cos(27π/16) + jsin(27π/16)).

So, the solutions to part (b) are the four complex numbers given above.