VerifyHow can I verify if the number (1+i) is the root of the equation z^4 + 4=0
You may also use the formula of difference of squares: a^2 - b^2 = (a-b)(a+b)
Writing the given equation as a difference of squares yields: z^4 - (-4) = 0 <=> (z^2)^2 - (2i)^2 = (z^2 - 2i)(z^2 + 2i) = 0
z^2 - 2i = 0 => z^2 = 2i
Writing the complex number z = x+iy => (x+iy)^2 = 2i
Expanding the binomial yields: x^2 + 2ixy - y^2 = 2i
Equating the coefficients of the same variable both sides yields:
x^2 - y^2 = 0
2xy = 2 => xy = 1 => x = y = 1
The number z is z = 1 + i.
This proves that z = 1 + i is a root of the equation z^4 + 4 = 0.
If a number is a root of an equation, that means that if substituting the number into the equation, it will verify the equation.
So, if the number (1+i) is the root of the equation
z^4 + 4=0, that means that (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 write it into it's trigonometric form, so that Moivre formula to be applied.
We'll write (1+i) into it's trigonometric form. We know that a complex number written into it's trig. 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
Just applying the Moivre formula, 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)
Best thing is to substitute 1+i for z in z^4+4 and get the value. If the value is 0 then (1+i) is a root of the equation, z^4+4 = 0 as below:
The value of the expression on the left side of the equation is
Expanding (1+i)^4 binimially we get:
The value of the left side of the equation for z =1+i is 0 and is equal to the value on the right.
So 1+i is root of z^4 + 4 = 0