# VerifyHow can I verify if the number (1+i) is the root of the equation z^4 + 4=0

*print*Print*list*Cite

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)

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

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

**(1+i)^4=-4 q.e.d.**

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

(1+i)^4+4.

Expanding (1+i)^4 binimially we get:

(1+4i^3+6i^2+4i+4)+4

1-4i-6+4i+4)+4

=0

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