Solve $z^5 +32 =0$

Question

Solve $z^5 +32 =0$

My attempt : $$z^5 = -32$$ Multiply the powers on both sides by $\frac{1}{5}$

we get $$z = 2 * (-1)^\frac{1}{5}$$

Now I'm stuck at this step I don't know how to proceed. Kindly help.

Hint: De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula — Justin Benfield, Jun 12 '16 at 10:40
If you want real-number solutions, use $(-1)^{1/5} = -1$. If you want complex solutions, see answer below. (But who knows, maybe you want numbers in some other system.) — GEdgar, Jul 22 '16 at 21:00

score 6 · Answer 1 · answered Jun 12 '16 at 10:30

Use polar coordinates:

$$z^5=32\cdot(-1)=2^5e^{\pi i+2k\pi i}=2^5e^{\pi i(1+2k)}\;,\;\;k\in\Bbb Z\implies$$

$$\implies z_k:=2\,e^{\frac{\pi i}5(1+2k)}\;,\;\;k=0,1,2,3,4$$

Check the above, and fill in details: why in the last line it is enough to take $\;k=0,1,2,3,4\;$ ? You may to check what you get with $\;k=5,6\;$ , say. Are all the $\;z_k$'s defined above different?

score 1 · Answer 2 · edited Apr 13 '17 at 12:20

1

We have $$\left(-\dfrac z2\right)^5=1$$

Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$

If $x^5-1=0,$

either $x-1=0\iff x=1$

or $x^4+x^3+x^2+x+1=0$

Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,

as $x\ne0,$ divide both sides by $x^2$ to find $$0=x^2+\dfrac1{x^2}+x+\dfrac1x+1=\left(x+\dfrac1x\right)^2+\left(x+\dfrac1x\right)-1$$

Solve for $x+\dfrac1x$ then solve for $x$

edited Apr 13 '17 at 12:20

Community

1

answered Jun 12 '16 at 11:36

lab bhattacharjee

274,582

Following this idea I get $x+\frac1x=\phi;,;;\phi=\frac{1+\sqrt5}2 =,$ the golden mean, and then $$x^2-\phi x+1=0\implies x_{1,2}=\frac{\phi\pm{\sqrt{\phi^2-4}}}2$$ but $$\phi^2-4=\frac{6+2\sqrt5}4-4=\frac{2\sqrt5-10}4=\frac{\sqrt5-5}2=-\sqrt5\frac{1-\sqrt5}2=\sqrt 5,,\phi^{-1}$$ I must be missing something: how does this lead to some specific result? Or is the idea to leave the solutions as functions of the golden mean? – DonAntonio Jun 12 '16 at 12:11
@Joanpemo, For $$x+\dfrac1x=\dfrac{\sqrt5+1}2,$$ $$x=\dfrac{\sqrt5+1}4\pm i\dfrac{\sqrt{10-2\sqrt5}}4$$ – lab bhattacharjee Jun 12 '16 at 13:10

score 1 · Answer 3 · edited Jun 12 '16 at 15:51

$z$ to the power of a fraction (let's define it as $1$ over $y$) is equal to the $y$th root of $z$.

Thus $-1$ to the power of $1/5$ is equal to the fifth root of $-1$, which is $-1$. Solving the equation now from this point is easy:

$$z = 2 \times (-1)^5 = 2 \times -1 = -2$$

Thus

$$z = -2$$

Check the answer by plugging it in:

$$z^5 + 32 = 0$$ $$-2^5 + 32 = 0$$ $$-32 + 32 = 0$$ $$0 = 0$$

score 0 · Answer 4 · answered Jun 12 '16 at 16:00

Since $(-1)^5=-1$, $(-1)^{\frac15}$ is equal to $-1$ in the real numbers realm. But consider $(\cos \frac{2k\pi}5 + i\sin \frac{2k\pi}5)^5=1$ so by multiplying the first answer you get all the answer as $$-2(\cos \frac{2k\pi}5 + i\sin \frac{2k\pi}5),\quad k=0, 1, 2, 3, 4$$

Solve $z^5 +32 =0$

4 Answers4