Finding modulus of complex number z in given equation

Question

If complex number $z$ satisfying

$6z^5+5iz^4+5iz-6=0$.

Then $|z|=$

I am trying to factor

$6z^5+5iz^4+5iz-6=0$

$6(z^5-1)+5iz(z^3+1)=0$

But I did not getting any common factor

I am also trying to factorise using

$i^2=-1$

$6z^5+5iz^4+5iz+6i^2=0$

$z^4(6z+5i)+i(5z+6i)=0$

I could not get any common factor

How do I find $z$ and then $|z|$

Please have a look. Thanks

FYI, your polynomial has a similar form to the one asked about in Modulus of complex root (found using an Approach0 search), with this answer there likely to be of the most use to you, along with the answers at the duplicate question it's closed of at If $11z^{10}+10iz^9+10iz-11 = 0$. Then possible value of $\mid z \mid,$ is. — John Omielan, Jan 11 '23 at 03:07
another way to solve is to notice that $z \ne 0$ so $\frac{1}{z^4}=-\frac{6z+5i}{5iz-6}=\frac{z+5i/6}{1+\overline{5i/6}z}$; but now it is an easy exercise to show that RHS preserves the unit circle (being a disc automorhism $\frac{z+a}{1+\bar a z}, a=5i/6, |a|<1$) and also its interior and exterior, while LHS preserves the unit circle but switches the interior with the exterior, so any such $z$ must perforce satisfy $|z|=1$ — Conrad, Jan 11 '23 at 04:14

score 1 · Answer 1 · answered Jan 11 '23 at 05:51

we substitute $z = r cis(θ)$ into the equation:

$6(r cis(θ))^5 + 5i(r cis(θ))^4 + 5i(r cis(θ)) - 6 = 0$

Next, we expand the powers:

$6r^5 cis(5θ) + 5ir^4 cis(4θ) + 5ir cis(θ) - 6 = 0$

Now, we equate the real and imaginary parts separately to zero.

$6r^5 cos(5θ) + 5ir^4 cos(4θ) = 0 and 5ir^4 sin(4θ) - 6 = 0$

Dividing the second equation by $5ir^4$ gives: $sin(4θ) = 6/5ir$.

From the first equation,

$6r^5 cos(5θ) = 0$, so either $r = 0$ or $cos(5θ) = 0$

And since we cannot have |r|=0, because it doesn't make sense for the magnitude, so we have cos(5θ) = 0.

This gives us $θ = π/5 + 2πn$ for some integer n.

Now, substituting this into the second equation, we get:

$6/5ir = sin(4θ) = sin(4(π/5 + 2πn)) = -sin(4π/5) = -0.3090169943749474$

so we have

$r^4 = -6/(5i*0.3090169943749474) = -1.9396926207859084$

and finally, the magnitude is given by the absolute value of this

$|z| = \sqrt{r^2} = |-1.9396926207859084| = 1.9396926207859084$

So the magnitude of the complex number z is approximately 1.94

Keep in mind that this is only one of the solutions of the equation, there are other solutions as well in the form of $cis(nπ/5 + 2πn)$ where n is an integer

Finding modulus of complex number z in given equation

1 Answers1