Prove that 1+i is one of the roots of z^4=-4.
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
= 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))
(1+i)^4=2*2*(cos pi + i* sin pi)
(1+i)^4=-4 q.e.d, so 1 + i is the root of the equation z^4 + 4=0.