Solving Equations of Complex Numbers

[ENGR_student]

Homework Statement

Show that (1+i) is a root of the equation z⁴=-4 and find the other roots in the form a+bi where (a) and (b) are real.

Homework Equations

Using De Moivre's Theorem
zⁿ=[rⁿ,nθ]
Modulus(absolute value of z) = 4
Argument = ?

The Attempt at a Solution

r⁴=4 → r = (4)^(1/5)
argument(z) = ∏/4 (not sure if that was right...)

Let:
[r⁴],5θ] = [4,2n∏ + ∏/4]

and then solve for the solutions for (n=-1,-2,1,2...im assuming to make it symmetrical)

 

[jackmell]

Homework Statement

Show that (1+i) is a root of the equation z⁴=-4 and find the other roots in the form a+bi where (a) and (b) are real.

Homework Equations

Using De Moivre's Theorem
zⁿ=[rⁿ,nθ]
Modulus(absolute value of z) = 4
Argument = ?

The Attempt at a Solution

r⁴=4 → r = (4)^(1/5)
argument(z) = ∏/4 (not sure if that was right...)

Let:
[r⁴],5θ] = [4,2n∏ + ∏/4]

and then solve for the solutions for (n=-1,-2,1,2...im assuming to make it symmetrical)

No.

Where is 1+i in the complex plane? You can draw it, then the absolute value is just \sqrt{2} and the principle argument, \theta, is \pi/4 right? So then 1+i=\sqrt{2} e^{\pi i/4}. Now what do you get when you raise that expression to the fourth power?

And in general, when you have an expression z^n=-k, we obtain z=(-k)^{1/n} and

(-k)^{1/n}=k^{1/n}e^{i/n(\pi+2j\pi)}, \quad j=0,1,2,\cdots,n-1

 

[HallsofIvy]

Personally, I wouldn't bother with "polar form". The first part asks you to "Show that (1+i) is a root of the equation z^4=-4". That, alone, does NOT ask you to solve the equation. You can "show that 'a' is a root of the equation f(x)= b" by evaluating f(a) and showing that it is 'b'. Here, if z= 1+ i then z^2= (1+ i)^2= 1^2+ 2(1)(i)+ i^2= 1+ 2i- 1= 2i. Then z^4= ((1+ i)^2)^2= (2i)^2= -4.

You should also know that "if a+ bi is a root of a polynomial equation with real coefficients then the complex conjugate, a- bi is also a root". Having determined at 1+i is a root, we immediately know that 1- i is also a root. Knowing that 1+ i is a root, we know that z- (1+ i) is a factor of the polynomial z^4+ 4 and knowing that 1- i is a root, we know that z- (1- i) is a factor. That is, we can write z^4+ 4= (z- (1+i))(z- (1- i))P(x).

But (z- (1+i))(z- (1- i))= ((z- 1)- i)((z- 1)+ i)= (z- 1)^2- i^2= z^2- 2z+ 1- i^2= z^2- 2z+ 2 so that z^4+ 4= (z^2- 2z+ 2)P(z). Dividing z^4+ 4 by z^2- 2z+ 2 gives P(z)= z^2+ 2z+ 2. Use the quadratic formula to solve P(z)= z^2+ 2z+ 2= 0.