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Solve the following system under the complex numbers (without eulerian form or polar form)

$$z^3 + w^5 = 0 \\ z^2 \bar w^4 = 1$$

I have found that $(\pm 1, \mp 1)$ satisfy the equations as well as letting $z,w$ be some powers of $i$ and using divisibility.

Are there any other solutions?

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Starting with the equations: $$z^3 + w^5 = 0 \\ z^2 \bar w^4 = 1$$ we find that $$w^4 = \bar z^{-2}.$$ Pluging this into the first equation yields $$z^3 + \bar{z}^{-2} w = 0$$

Which is equivalent to: $$|z|^4 z = - w$$ by noting that $z\cdot \bar z = |z|^2$.

Finally replacing this back into the original equation: $$z^2 \bar w^4 = z^2 \bar z^4 |z|^{16} = \bar z^2 |z|^{20} = 1$$

This means that $|z|^{21}=1$, and $|z| = 1$. Hence $\bar z^2 = 1$, and $w=-z$. There are only two roots to the equation $\bar z^2 - 1$, so $z$ could be $\pm 1$. Also $w = -z$. This gives the solutions $(1,-1)$ and $(-1,1)$. Which can be seen to be valid.

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  • $\begingroup$ You've missed square in $\bar z^2 |z|^{20}$ $\endgroup$ – uranix May 7 '15 at 20:24
  • $\begingroup$ So I did... I will edit it. $\endgroup$ – Joel May 7 '15 at 20:25
  • $\begingroup$ Thank you for catching the mistake @uranix $\endgroup$ – Joel May 7 '15 at 20:28
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By taking absolute values of $z$ and $w$ from the first equation $$|z|^3 = |w|^5$$ Since neither $z$ nor $w$ can be zero, $$|z| = |w|^{5/3}.$$ From the second $$1 = |z|^2|w|^4 = |w|^{22/3},$$ so $$|w| = |z| = 1.$$ Since $|w| = 1$, $\bar{w} = w^{-1}$ and $z^2 = w^4$. Substituting in the first equation gives $$w^4(z + w) = 0,$$ so $z = -w$. Plugging that into $w^4 = z^2$ gives $$w^4 = w^2 \Leftrightarrow w^2(w^2 - 1) = 0.$$ So there are no solutions except $(\pm 1, \mp 1)$.

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  • $\begingroup$ Why there are no other solutions? When we are taking absolute values we might miss some solutions since it is not an $\iff$. $\endgroup$ – bolzano May 7 '15 at 20:13
  • $\begingroup$ No, we can introduce new solutions, not miss. Consider $x = 1 \Rightarrow |x| = |1| = 1, x = \pm 1$. $\endgroup$ – uranix May 7 '15 at 20:14
  • $\begingroup$ Its equivalent to the difference between finding the curve or the geometric locus (may include more points) on a given equation $\endgroup$ – bolzano May 7 '15 at 20:15
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Let $z=re^{it}$. The second equation then gives us $$\bar{w}^4 = r^{-2}e^{-2it} \implies \bar{w} = r^{-1/2}e^{-i(t+n\pi)/2} \implies w = r^{-1/2}e^{i(t+n\pi)/2}$$ We also need $$z^3 + w^5 = 0 \implies r^3e^{3it} = -r^{-5/2}e^{5i(t+n\pi)/2} \implies r^{11/2}e^{3it-5it/2-5n\pi i/2} = 1$$ This gives us $r=1$ and $\left(t/2-5n\pi/2\right) = 2k\pi \implies t = 5n\pi+4k\pi$. Hence, $z = \pm1$. Hence, the only solutions are $$(z,w) = \pm (1,-1)$$

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  • $\begingroup$ Thanks I a solution with the use of algebraic manipulations and the cartesian form (not $re^{it}$). Anyway thanks $\endgroup$ – bolzano May 7 '15 at 20:11

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