2

I asked this question here Solving $z^4 + 4i\bar{z} = 0$

Though I accepted the answer, later I found out that I and WA don't agree on the result.

What I did:

Find the absolute value of $z$ by solving $|z^4| = |-4i\bar{z}|$

Getting $|z|^4 = 4|z|$ and hence $|z| = 0$ or $|z| = 4^{\frac{1}{3}}$

Leaving the trivial solution $z=0$ aside, we have $|z| = 4^{\frac{1}{3}}$

Now back to the original equation, multiplying it by $z$ we get:

$z^5 = -4i|z|^2$ and hence $z^5 = -4i4^{\frac{2}{3}}$ hence $z^5 = 4^{\frac{5}{3}}\operatorname{cis}(-\frac{\pi}{2})$

Getting the roots $z_k = 4^{\frac{1}{3}}\operatorname{cis}(\frac{-\frac{\pi}{2} + 2\pi k}{5})$ where $k=0,1,2,3,4$

So we get the multiplication of the complex roots to be:

$z_0\cdot z_1\cdot z_2\cdot z_3\cdot z_4 = 4^{\frac{5}{3}}\operatorname{cis}(\frac{-\pi}{10} + \frac{3\pi}{10} + \frac{7\pi}{10} + \frac{11\pi}{10} + \frac{15\pi}{10}) = 4^{\frac{5}{3}}\operatorname{cis}(\frac{35\pi}{10}) = -4^{\frac{5}{3}}i=-2^{\frac{10}{3}}i$

But when calculating with WA I get $-2^{\frac{5}{2}}i$.

What is the correct result?

Aladin
  • 702
  • 2
    Do you see that $35\pi/10 = 7\pi/2$? – B. Goddard Feb 19 '21 at 14:18
  • Yes I edited the answer to add that calculation but it still differs from the computer answer – Aladin Feb 19 '21 at 14:21
  • Hint: Let $z=re^{i\varphi}$. Now compare the phase angle of $z$ and $z^*$. – vitamin d Feb 19 '21 at 14:23
  • Letting $z=re^{i\theta}$ gives $$r^4e^{(5\theta)i}=4(-i)=4re^{(2k\pi+3\pi/2)i}$$ – Mikasa Feb 19 '21 at 14:28
  • Did the computer really get the five halves power of $2$? – Barry Cipra Feb 19 '21 at 14:36
  • Yes, Wolfram alpha to be exact, moreover this question appears in a past test (a multiple choice question) and the Wolfram alpha answer appears as one of the answers and my answer does not appear. – Aladin Feb 19 '21 at 14:45
  • @Aladin, huh! Something fishy is going on, It might help if you included a link to the WA calculation that gives the different answer, so we can see if there's some discrepancy in your input there. (It might also help to clarify at the beginning of your post here that what you're really asking about is the product of the nonzero roots of the equation $z^4+4i\overline{z}$, and why you and WA disagree.) – Barry Cipra Feb 19 '21 at 15:04
  • I put it in WA and there are cube roots galore: https://www.wolframalpha.com/input/?i=z%5E4%2B4iConjugate%5Bz%5D%3D0 – B. Goddard Feb 19 '21 at 15:22
  • I took the plain text of the roots and multiplied them. It didn't allow me to put the total product since it was too long so had to run four (equivalent) queries and this was the final result. – Aladin Feb 19 '21 at 18:01
  • Anyway, is my result correct? – Aladin Feb 21 '21 at 09:41

1 Answers1

1

Wolframalpha returns the product of the roots of $\displaystyle\frac{z^4 + 4 i z^*}z = 0$ as $$-10.0794 i,$$ which indeed actually equals the $$-2^{\frac{10}{3}}i$$ that you obtained.

(First dividing the equation by $z$ was to discard the trivial root $z=0,$ as desired.)

ryang
  • 38,879
  • 14
  • 81
  • 179