2 Answers | Add Yours
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
This proves that 1 + i is a root of z^4=-4.
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.
We’ve answered 319,651 questions. We can answer yours, too.Ask a question