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I'm trying to find the complex roots of $z^3 + \bar{z} = 0$ using De Moivre.

Some suggested multiplying both sides by z first, but that seems wrong to me as it would add a root ( and I wouldn't know which root was the extra ).

I noticed that $z=a+bi$ and there exists $\theta$ such that the trigonometric representation of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$ .

It seems that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$

However, my trig is pretty rusty and I'm not quite sure where to go from here.

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It doesn't add a root to multiply by $z$ since $z=0$ is already a root. –  Thomas Andrews May 14 '12 at 18:04
    
Why is $z=0$ already a root? –  Robert S. Barnes May 14 '12 at 18:22
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Another option, a sort of brute-force technique, is to write $z=a+bi$ and then equate real parts and imaginary parts to get two third-degree equations in $a$ and $b$. This sounds scary, but since you already know that $0+0i$ is a root, you will easily be able to eliminate this solution from the third-degree equations to get two second-degree equations in $a$ and $b$ which you should be able to solve. Let me know if you would like to see this written up in more detail. –  MJD May 14 '12 at 18:35
    
Another thing that might be helpful in such problems: if $z\in\Bbb C$ and $|z|=1$ then $\overline{z}=z^{-1}$. –  Dejan Govc May 14 '12 at 19:27
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@RobertS.Barnes What is $0^3 + \bar 0$? –  Thomas Andrews May 14 '12 at 19:30
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6 Answers 6

Hint: First try to narrow down the value of $|z|$

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As has been observed by Thomas Andrews, $z=0$ is a root, so we will not be introducing extraneous roots if we multiply by $z$, or by $\overline{z}$.

From the answer by Aryabhata, you should be able to conclude that if $z\ne 0$, then $z$ has norm $1$.

Now it may be simplest to multiply by $\overline{z}$. We get the pleasantly symmetrical equation $z^2+\overline{z}\overline{z}=0$. If $z$ has norm $1$, let $z=\cos\theta+i\sin\theta$. Then use De Moivre's Theorem.

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What does $z$ has norm 1 mean? –  Robert S. Barnes May 14 '12 at 18:21
    
The norm of $z=a+ib$ is $\sqrt{a^2+b^2}$. Equivalently, it is $\sqrt{z\overline{z}}$. –  André Nicolas May 14 '12 at 18:22
    
Ahh, I'm used to calling it absolute value, $|z|$. If I multiply by $\bar{z}$ I get $z^4 = -|z|^2$. How do I conclude from that, that $|z|=1$? –  Robert S. Barnes May 14 '12 at 18:37
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Norms preserve multiplication. That is, if $c=ab$ then $|c| = |a|\cdot|b|$. Let's refer to the norm of $z$ as $n$, for "norm". You have $z^3 = -\bar z$, so $|z^3| = |-\bar z|$. The right-hand side goes like this: $|-\bar z| = |z| = n$. (That's an exercise for you.) And $|z^3| = {|z|}^3 = n^3$. Since $z^3 = -\bar z$, their norms must be equal and we have $n^3 = n$. The norm $n$ is always a non-negative real number, so $n=0$ or $n=1$ are the only possibilities. –  MJD May 14 '12 at 18:41
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The norm of a product is the product of the norms. So from $z^3=-\overline{z}$ you conclude that $|z|^3=|-\overline{z}|=|z|$. Therefore $|z|=0$ or $|z|=1$. In case $|z|=1$, multiply by $\overline{z}$. We get $z^2|z|^2+\overline{z}\overline{z}=0$. –  André Nicolas May 14 '12 at 18:42
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up vote 1 down vote accepted

OK, I'm going to take a stab at this.

Given: $z^3 + \bar{z} = 0$

Therefore $z^3 = -\bar{z}$ and $|z^3| = |-\bar{z}|$ .

We have that $|-\bar{z}|=|\bar{z}|=|z|$ therefore $|z^3|=|z|$ and $|z||z||z|=|z|$.

Therefore $|z|=0$ or $|z|=1$.

Case 1: Assume $|z|=0$ then if

$z=a+bi\leftrightarrow|z|=\sqrt{a^2+b^2}=0\leftrightarrow a=0 \wedge b=0\leftrightarrow a+bi=0$

thus $z = 0$.

Case 2: Assume $|z|=1$

Let's multiply by $z$ and we get $z^4 = -z\bar{z}$. We see that $z\bar{z} = |z|^2$ so we get $z^4 = -( |z|^2 )$ so from the above either:

therefore $z^4 = -|z| = -1$

The trigonometric representation of $-1$ is $1*( \cos \pi + i \sin \pi )$ so according to De Moivre:

$z^4 = r^4(\cos 4\theta + i \sin 4\theta ) = 1*( \cos \pi + i \sin \pi )$

These are two complex numbers in trigonometric form so:

$r^4 = 1$ and $4\theta = \pi + 2\pi*k$ or

$r=1$ and $\theta = \frac{\pi + 2\pi*k}{4}$ and each solution has the form:

$z_k = \cos( \frac{\pi + 2\pi*k}{4} ) + i \sin (\frac{\pi + 2\pi*k}{4})$

for $0\leq k \leq 3$.

Which together with Case 1 gives the following values for $z$:

$0,\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$

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Can someone check my answer? Thanks! –  Robert S. Barnes May 15 '12 at 15:13
    
This is correct. And as Zarrax noted, there are 4 possibilities for $z_k$ –  M Turgeon May 15 '12 at 15:21
    
$\pm\frac{1}{\sqrt{2}}$ does not have absolute value 1, and so cannot be a solution, as you proved. –  M Turgeon May 15 '12 at 17:55
    
I must have screwed something up typing it into wolfram. I typed: solve \cos ( ( \pi+0*2*\pi )\4 ) + i*\sin ( \pi+0*2*\pi )\4 ) into wolphramalpha and it gave me $\frac{1}{\sqrt{2}}$ –  Robert S. Barnes May 15 '12 at 17:57
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Write $z = re^{i\theta}$. Then you are trying to solve $r^3e^{3i\theta} + re^{-i\theta} = 0$, which is the same as $$r^3e^{3i\theta} = -re^{-i\theta}$$ Note that $-1 = e^{i\pi}$, so the above is equivalent to $$r^3e^{3i\theta} = re^{i(\pi - \theta)}$$ Comparing magnitudes, you have $r^3 = r$, which is solved by $r = 0$ and $1$, and comparing arguments you must have $3\theta = \pi - \theta + 2\pi k$ for some integer $k$ (when $r \neq 0$). Thus for some integer $k$ you have $$\theta = {\pi \over 4} + k{\pi \over 2}$$ There are four values of $\theta$ in $[0,2\pi)$ that satisfy this, namely ${\pi \over 4}, {3\pi \over 4}, {5\pi \over 4}$, and ${7\pi \over 4}$. Thus the complex numbers satisfying your original equation are $0, e^{i {\pi \over 4}}, e^{i {3\pi \over 4}}, e^{i {5\pi \over 4}}$, and $e^{i {7\pi \over 4}}$. In rectangular coordinates these are $0$ and $\pm {1 \over \sqrt{2}} \pm {i \over \sqrt{2}}$.

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I got 0 and $\pm\frac{1}{\sqrt{2}}$ as the answers. Maybe you could check against my solution and see what's correct? –  Robert S. Barnes May 15 '12 at 17:50
    
Your solution is right until right before the end... you left out the imaginary parts of the solutions, $\sin({\pi \over 4}) = {1 \over \sqrt{2}}$ etc. –  Zarrax May 15 '12 at 18:03
    
I miss typed into wolfram - () weren't balanced. –  Robert S. Barnes May 15 '12 at 18:09
    
I'm still getting strange stuff in wolfram: $\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$ –  Robert S. Barnes May 15 '12 at 18:16
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The wolfram solutions mean what I wrote. Just take sine and cosine of the various arguments and you'll see the answer. It can have 5 solutions... only polynomials of degree 3 will have to have at most 3 solutions and this isn't a polynomial. –  Zarrax May 15 '12 at 18:37
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EDIT in view of the comments bellow by JimConant and PeterTaylor. If there is still any error the fault is mine.

This is an alternative solution to the trigonometric one. We will use the algebraic method. Let $z=x+iy$. We have $$\begin{eqnarray*} 0 &=&z^{3}+\overline{z} \\ 0 &=&\left( x+iy\right) ^{3}+\left( x-iy\right) \\ &=&x^{3}+x-3xy^{2}+i\left( 3x^{2}y-y^{3}-y\right) \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x(x^{2}+1-3y^{2}) \\ 0=y(y^{2}+1-3x^{2}). \end{array} \right. \end{eqnarray*}\tag{1}$$

One of the roots is $$x_{1}=y_{1}=0.\tag{1a}$$ The remaining real roots satisfy the system

$$\begin{eqnarray*} \left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=y^{2}+1-3x^{2} \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=\frac{1}{3}+\frac{1}{3}x^{2}+1-3x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=4-8x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=1-2y^{2} \\ 0=1-2x^{2}. \end{array} \right. \end{eqnarray*} \tag{2}$$

The last system means that $$y=\pm x\tag{3}$$ and that

$$\begin{eqnarray*} x &=&\pm\frac{1}{2}\sqrt{2}, \\ y &=&\pm\frac{1}{2}\sqrt{2}. \end{eqnarray*}\tag{3a}$$

Combining the above results, we conclude that the following five complex numbers

$$ z_{1} =0,\tag{4}$$ $$ z_{2} =\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2},\quad z_{3} =-\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2}, \tag{5}$$ $$z_{4} =\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2},\quad z_{5} =-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2}, \tag{6}$$

are the solutions of the given equation $$z^{3}+\overline{z}=0.\tag{7}$$

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Is it really true that having equations symmetric with respect to $x$ and $y$ implies $x=y$? It seems to me this only implies that if $(x,y)$ is a solution then so is $(y,x)$. –  Grumpy Parsnip May 15 '12 at 18:45
    
@JimConant: You are right! I was wrong. –  Américo Tavares May 15 '12 at 18:50
    
@JimConant Thanks for finding the error. I will try to reformulate the solution. –  Américo Tavares May 15 '12 at 19:27
    
@JimConant I edited the answer. I hope now the solution is correctly justified. –  Américo Tavares May 15 '12 at 19:57
    
You only find 3 solutions, but you haven't explained why you're discarding two (x, y) pairs. –  Peter Taylor May 15 '12 at 20:01
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That is a comment to the comment "should there be only 3 roots?" in Zarrax answer. Actually, the questions is "why isn´t there 9 roots?". This is the right question to ask since the intersection of the two curves in $\mathbb{R}^2$ $$ x^3-3xy^2 +x = 0$$ $$ 3x^2y-y^3 -y = 0$$ is your solution set (just expand out $z^3+\bar{z}$). Now, the intersection of two degree 3 equations should have $3x3= 9$ solutions (by Bezout´s theorem). The 4 roots we are missing at "infinity" or in the $\mathbb{C}^2$ plane. We could apply the same thing to the function $z^2+z$. As complex polynomial, we should expect 2 roots. As the real system $(x,y) \rightarrow (x^2-y^2+x,2xy+y)$ we expect four real roots.

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According to my computations the 4 "missing" roots are in the $\mathbb{C}^2$ plane: $(0,i),(0,-i),(i,0),(0,i)$. –  Américo Tavares May 15 '12 at 20:51
    
Correction: $(0,i),(0,−i),(i,0),(-i,0)$. –  Américo Tavares May 15 '12 at 22:02
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