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

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is given that z^4 = -4. To prove that 1 + i is one of the roots, substitute z with 1 + i

=> (1 + i)^4

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

use i^2 = -1

=> (2i)^2

=> 4*i^2

=> -4

This proves that 1 + i is a root of z^4=-4.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 318,944 questions. We can answer yours, too.

Ask a question