# Prove that 1+i is one of the roots of z^4=-4.

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Given the function z^4 = -4

We need to determine if 1+ i is a root of z^4 = -4.

Let us substitute;

(1 + i )^4 = -4

==> Let us rewrite the power.

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

But we know that i = sqrt-1 ==> i^2 = -1.

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

= (2i)^2

= (4*i^2)

= 4*-1 = -4

Then we proved that if z^4 = -4.

Then **(1+i) is one of the roots** of the equation.

To prove that a number is a root of an equation, we'll have to prove that substituted the number into the equation, it will cancel the equation.

Let's check if the number (1+i) is the root of the equation, namely if it cancels the equation z^4 + 4=0.

(1+i)^4 + 4=0

So, (1+i)^4 = -4

We know that in order to raise a complex number to a power, we'll have to put the number in polar form, so that Moivre's formula to be applied.

We'll write (1+i) into it's polar form.

We know that a complex number written into it's polar form, has the following form:

z=r*(cos t+ i*sin t), where r=sqrt(a^2+b^2) and tg t= (b/a)

So, for the number 1+i, r=sqrt(1^2 + 1^2)=sqrt 2

tg t= 1/1=1, so t= pi/4

1+i=sqrt 2*(cos (pi/4) + sin(pi/4))

(1+i)^4=(sqrt 2)^4*(cos (pi/4) + i*sin(pi/4))^4

We'll apply Moivre's formula and we'll have:

(1+i)^4=(sqrt 2)^4*(cos 4*(pi/4) + i*sin4*(pi/4))

We'll simplify:

(1+i)^4=2*2*(cos pi + i* sin pi)

(1+i)^4=2*2*(-1+ i*0)

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

(1+i)^4=-4 q.e.d, **so 1 + i is the root **of the equation z^4 + 4=0.