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

chwala
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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 ...

1665319812803.png
 
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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
 
A systematic approach is to factorize <br /> 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: <br /> \begin{split}<br /> z^3 \quad &amp;: \quad A + B = 0 \\<br /> z^2 \quad &amp;: \quad 8 + AB = -1 \\<br /> z \quad &amp;: \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).
 
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chwala said:
How did they come to

completion of the square
 
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malawi_glenn said:
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##
 
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chwala said:
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 ##
 
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malawi_glenn said:
##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!
 
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:$$-b = -b\frac{|w|^2}{16} \Rightarrow |w|^2 = 16 \Rightarrow a^2 + b^2 = 16$$$$a = 1-a \Rightarrow a = \frac 1 2 \Rightarrow b^2 = -\frac{63}{4}$$$$\Rightarrow z^2 = \frac 1 2 \pm \frac{3\sqrt 7}{2}i$$
 
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chwala said:
@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
 
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