complex number question (1 Viewer)

cormglakes · Mar 19, 2021

let z=cosx + isinx

solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0

Qeru · Mar 19, 2021

cormglakes said:
let z=cosx + isinx

solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0

$z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

$z^4+1 - 4(z^3+z) + 2z^2 = 0$

$z^2+\frac{1}{z^2}-4 \left (z+\frac{1}{z} \right) +2=0 \quad \text{dividing by $z^2$}$

But $z^n+\frac{1}{z^n}=2\cos{nx}$

So the problem becomes: $2\cos{2x}-8\cos{x}+2=0$

which is a 3U problem (remember to take only the first 4 unique solutions).

OR

notice $z^2+\frac{1}{z^2}=\left (z+\frac{1}{z}\right)^2-2$

So the problem becomes: $\left (z+\frac{1}{z}\right)^2-4\left(z+\frac{1}{z}\right)=0$

Which is simply a quadratic in $z+\frac{1}{z}$

cormglakes · Mar 19, 2021

Qeru said:
$z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

$z^4+1 - 4(z^3+z) + 2z^2 = 0$

$z^2+\frac{1}{z^2}-4 \left (z+\frac{1}{z} \right) +2=0 \quad \text{dividing by $z^2$}$

But $z^n+\frac{1}{z^n}=2\cos{nx}$

So the problem becomes: $2\cos{2x}-8\cos{x}+2=0$

which is a 3U problem (remember to take only the first 4 unique solutions).

and then after I get cosx=0 --> z= i & -i
and cosx=2 --> how do I find the other 2 roots?

CM_Tutor · Mar 19, 2021

Qeru said:
$z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

$z^4+1 - 4(z^3+z) + 2z^2 = 0$

At this point, notice we have a quadratic in $z^2$ ? So...

$\begin{align*} z^4 + 2z^2 + 1 - 4z(z^2 + 1) &= 0 \\ (z^2 + 1)^2 -4z(z^2 + 1) &= 0 \\ (z^2 + 1)(z^2 + 1 - 4z) &= 0 \\ z^2 &= -1 \qquad \qquad \text{or} \qquad \qquad z^2 - 4z + 4 = 3 \\ z &= \pm i \qquad \qquad \text{or} \qquad \qquad z - 2 = \pm\sqrt{3} \end{align*}$

and the equation has two real roots, $z = 2 \pm \sqrt{3}$ , and two purely imaginary roots $z = \pm i$

CM_Tutor · Mar 19, 2021

cormglakes said:
and then after I get cosx=0 --> z= i & -i
and cosx=2 --> how do I find the other 2 roots?

In taking $z = \cos{x} + i\sin{x}$ there is an assumption made that all solutions in $z$ will have modulus of 1 and satisfy $|z| = 1$ . This is not true for the two real solutions so you run into a problem.

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