Solve ##z^2(1-z^2)=16## using Complex numbers

[chwala]

Homework Statement

See attached

Relevant Equations

Complex numbers

The problem is as shown...all steps are pretty easy to follow. I need help on the highlighted part in red. How did they come to;
$z^4+8z^2+16-9z^2=0$ or is it by manipulating $-z^2= 8z^2-9z^2?$ trial and error ...

 

[anuttarasammyak]

The idea is to make formula of
$$A^2=B^2$$
. Let us see.
$$z^4+16=z^2$$
$$(z^2+4)^2=9z^2=(3z)^2$$
$$z^2+4=\pm 3z$$

 

[pasmith]

A systematic approach is to factorize $$z^4 - z^2 + 16 = (z^2 + Az + 4)(z^2 + Bz + 4)$$ and compare coefficients of powers of $z$ to get three equations for $A$ and $B$: $$\begin{split}<br /> z^3 \quad &: \quad A + B = 0 \\<br /> z^2 \quad &: \quad 8 + AB = -1 \\<br /> z \quad &: \quad 4A + 4B = 0<br /> \end{split}$$ Since the first and third are really the same equation, we can solve this system to obtain $(A, B) = (3, -3)$.

 

[malawi_glenn]

How did they come to

completion of the square

 

[chwala]

completion of the square

How did they come to $z^4+8z^2+16-9z^2$?...of course the step after is complete square method...no problem there. Or it was just simply identifying that;

$-9z^2+8z^2=-z^2$

 

[malawi_glenn]

How did they come to $z^4+8z^2+16-9z^2$?

$z^4-z^2+16 = z^4-z^2+16 + 0 = z^4-z^2+16 +8z^2 - 8z^2 = z^4+8z^2+16-9z^2$

 

[chwala]

$z^4-z^2+16 = z^4-z^2+16 + 0 = z^4-z^2+16 +8z^2 - 8z^2 = z^4+8z^2+16-9z^2$

@malawi_glenn smart move there mate!

 

[PeroK]

Here's another approach. Let $w = z^2$:$$ww^* = |w|^2 = 16\frac{|w|^2}{16} = w(1-w)\frac{|w|^2}{16}$$ $$\Rightarrow w^* = (1-w)\frac{|w|^2}{16}$$Comparing the real and imaginary parts with $w = a + ib$ gives:$$-b = -b\frac{|w|^2}{16} \Rightarrow |w|^2 = 16 \Rightarrow a^2 + b^2 = 16$$$$a = 1-a \Rightarrow a = \frac 1 2 \Rightarrow b^2 = -\frac{63}{4}$$$$\Rightarrow z^2 = \frac 1 2 \pm \frac{3\sqrt 7}{2}i$$

 

[malawi_glenn]

@malawi_glenn smart move there mate!

And you do this in order to complete the square adding and subtracting the same thing so it fits