# Solving Equations of Complex Numbers

1. Oct 12, 2013

### ENGR_student

1. The problem statement, all variables and given/known data
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.

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

3. 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)

2. Oct 12, 2013

### jackmell

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

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

3. Oct 12, 2013

### 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 $P(z)= z^2+ 2z+ 2= 0$.

Last edited by a moderator: Oct 12, 2013