Demonstrate that the complex number 1+i is the solution of equation z^4+4=0.

To show that 1+i is the solution of z^4+1 =
0.

We can either solve the equation, or we substitute for z
= (1+i) in z^4+4 and show that 1+i satisfies the
equation.

We know z^4 +4 = 0. So z^4 =
-4.

Z^4 = 4(-1+i*0)

Z^4 =
4{cos(2n+1)pi + i sin(2n+1)pi}

We take the 4th root in
accordance with D'Moivre's theorem. The theorem states that (cosx + i sinx)^n = (cosnx +
i sinnx) for all real n.

Z= 4^(1/4){cos(2n+1)pi
+isin(2n+1)pi}^(1/4).

z = 2^(1/2) {cos (2n+1)pi/4
+isin(2n+1)pi/4} for n = 0,1,2,3.

When n = 0, z = 2^(1/2)
{cos pi/4 +isinpi/40 = 2^(1/2){ 1/sqrt2+i1/sqrt2) =
1+i.

Therefore 1+i is the root of z^4+4 =
0.