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Solve $$z^5+4\overline z^3=0$$

This is what I did. Let $z=r(\cos\theta+i\sin\theta)$ so, the equation is: $$r^5(\cos5\theta+i\sin5\theta)+4r^3(\cos3\theta-i\sin3\theta)=0.$$ Suppose $r$ is not $0$ and divide both sides by $r^3$ to get: $$(r^2\cos5\theta+4\cos3\theta)+i(r^2\sin5\theta-4\sin3\theta)=0+0i$$ so we get the following system of equations: $$r^2\cos5\theta+4\cos3\theta=0$$ $$r^2\sin5\theta-4\sin3\theta=0$$ I'm not sure how to continue. I've tried subtract $4\cos3\theta$ from both sides in the first equation and in the second one add $4\sin3\theta$ to both sides, and then divide the equations, to get $\tan5\theta=-\tan3\theta$, but then i suppose im not dividing be zero and that's a problem. What can I do from here? Is there a simpler way to solve this problem? Thanks!

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3 Answers 3

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First of all, this is not polynomial. It looks like it, but it's not: the fundamental theorem of algebra doesn't apply (notice that we get nine solutions at the end, not five).

Notice that $z = 0$ is an obvious solution, so we will look for non-zero solutions. Multiply the equation by $z^3$ to get $$z^8 + 4|z^6| = 0\\ z^8 = -4|z|^6$$

so $z^8$ is negative real number, i.e. $8\arg z \equiv \pi \pmod {2\pi}$, by de Moivre's formula, and $\arg z = \frac{\pi + 2k\pi}{8}.$

On the other hand, $$z^8 = - 4|z|^6 \implies |z|^8 = 4|z|^6 \implies |z|^2 = 4 \implies |z|=2.$$

Thus, the solutions are $2e^{i\frac{\pi + 2k\pi}{8}},\ k\in\mathbb Z$ and $0$.

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  • $\begingroup$ Question based on my complete ignorance - why isn't it a polynomial? Clearly a hole in my basic knowledge - suggestions to fill it welcome. $\endgroup$
    – astaines
    Commented Jun 3, 2018 at 21:42
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    $\begingroup$ @astaines, to be more precise, it is not a polynomial in one variable $z$. If it were, $\bar z$ would be polynomial in $z$ as well and it is not (why?). Secondly, $z^5+4\overline z^3$ is not differentiable. Finally, as I've written, the fundamental theorem of algebra doesn't apply. If $z^5+4\overline z^3$ were polynomial, it would have to be equal to $(z-z_1)\cdots(z-z_k)$, where $z_i$ are the nine solutions that we've found. $\endgroup$
    – Ennar
    Commented Jun 3, 2018 at 21:58
  • $\begingroup$ Ok I begin to see - thank you! $\endgroup$
    – astaines
    Commented Jun 5, 2018 at 8:52
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Note that as per Euler's Formula the complex number $z$ can be expressed as $z=re^{i\theta}$ where $|z|=r$ and $\theta$ is the argument of $z$. Likewise, the complex number $\bar z $ can be expressed as $\bar z=re^{-i\theta}$ where $|\bar z|=r$ and $-\theta$ is the argument of $\bar z$.

Hence, you are looking for solutions to $$r^5e^{5i\theta}+4r^3e^{-3i\theta}=0$$

Two "solution branches" are possible: $r=0$ and $r\ne0$ $$$$ For $r=0$ the solution is obviously the number $0$ $$$$ For $r\ne 0$,

$$r^5e^{5i\theta}+4r^3e^{-3i\theta}=0$$ Dividing by $r^3$ (since $r\ne 0$) $$r^2e^{5i\theta}+4e^{-3i\theta}=0$$ $$\Rightarrow r^2e^{8i\theta}+4e^{0i}=0$$

$$\Rightarrow r^2e^{8i\theta}=-4e^{0i}=4e^{i\pi}=4e^{i(\pi+2k\pi)}$$ $$\Rightarrow r^2e^{8i\theta}=4e^{i(\pi+2k\pi)}$$ $$\Rightarrow r=2,\theta=\left(\dfrac{\pi}{8}+\dfrac{2k\pi}{8}\right)$$ where $k=0,1,2,3,...,7$

Note that $-1=e^{i\pi}$ $$$$

Thus all the solutions to the equation are $$z=0 , 2e^{i(\frac\pi8+\frac{2k\pi}8)}$$ where $k=0,1,2,3,...,7$

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Let's try pairing the given equation with its conjugate:

$z^5=4\overline{z}^3$

$\overline{z}^5=4z^3$

If we square the first equation we get

$z^{10}=16\overline{z}^6$

And then

$z^{10}=16\overline{z}^6=16\overline{z}^5\overline{z}$

$z^{10}=64z^3\overline{z}$

So either $z=0$ or else

$\overline{z}=z^7/64$

Now for the nonzero solutions we can go back to our original equation:

$z^5=4\overline{z}^3=4z^{21}/2^{18}$

$z^{16}=2^{16}$

and we get the solution set $\{0,2\exp[i(n\pi/8)]\}$.

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