The solutions of the equation z^2(1-z^2)=16,z | vedclass

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(D) Given equation is $z^2(1-z^2)=16$,where $z \in \mathbb{C}$.This can be rewritten as $z^2 - z^4 = 16$,or $z^4 - z^2 + 16 = 0$.Let $z^2 = w$. Then $

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$|z|=2$

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(D) Given equation is $z^2(1-z^2)=16$,where $z \in \mathbb{C}$.This can be rewritten as $z^2 - z^4 = 16$,or $z^4 - z^2 + 16 = 0$.Let $z^2 = w$. Then $w^2 - w + 16 = 0$.Using the quadratic formula,$w = \frac{1 \pm \sqrt{1 - 64}}{2} = \frac{1 \pm i\sqrt{63}}{2} = \frac{1 \pm 3i\sqrt{7}}{2}$.Since $w = z^2$,we have $|w| = |z^2| = |z|^2$.Calculating the modulus of $w$:$|w| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{3\sqrt{7}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{63}{4}} = \sqrt{\frac{64}{4}} = \sqrt{16} = 4$.Therefore,$|z|^2 = 4$,which implies $|z| = 2$.

The solutions of the equation $z^2(1-z^2)=16$,$z \in \mathbb{C}$,lie on the curve

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