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

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If 1 + i is a solution of z^44 + 4 = 0, if we substitute z with 1 + i, we should get 0.

Doing the same

(1 + i)^4 + 4

=> [(1 + i)^2]^2 + 4

=> [1 + i^2 + 2i]^2 +4

=> [1 - 1 + 2i]^2 + 4

=> (2*i)^2 + 4

=> 4* i^2 + 4

=> -4 + 4

=> 0

So we arrive at 0.

**Therefore we can say that 1 + i is a solution of z^4 + 4 = 0.**

The complex number is the root of the given equation if and only if substituted into equation, it verifies the equation.

We'll substitute z by 1+i:

(1+i)^4 + 4 = 0

First, we'll raise the brackets to square:

(1+i)^2 = 1 + 2i + i^2

We'll recall that i^2 = -1, in the complex number theory:

(1+i)^2 = 1 + 2i - 1

We'll eliminate like terms:

(1+i)^2 = 2i

Now, we'll raise to square the result:

[(1+i)^2]^2 = (2i)^2

(1+i)^4 = 4i^2

(1+i)^4 = -4

Now, we'll substitute the value of (1+i)^4 into the equation:

-4 + 4 = 0

**The number 1 + i has verified the given equation, so the complex number 1 + i is the root of the 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.