# complex number question (1 Viewer)

#### cormglakes

##### New Member
let z=cosx + isinx

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

#### Qeru

##### Well-Known Member
let z=cosx + isinx

solve z^4 - 4z^3 + 2z^2 - 4z+1 = 0
$\bg_white z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

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

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

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

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

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

OR

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

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

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

#### cormglakes

##### New Member
$\bg_white z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

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

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

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

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

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

##### Moderator
Moderator
$\bg_white z^4 - 4z^3 + 2z^2 - 4z+1 = 0$

$\bg_white z^4+1 - 4(z^3+z) + 2z^2 = 0$
At this point, notice we have a quadratic in $\bg_white z^2$? So...

\bg_white \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, $\bg_white z = 2 \pm \sqrt{3}$, and two purely imaginary roots $\bg_white z = \pm i$

#### CM_Tutor

##### Moderator
Moderator
and then after I get cosx=0 --> z= i & -i
and cosx=2 --> how do I find the other 2 roots?
In taking $\bg_white z = \cos{x} + i\sin{x}$ there is an assumption made that all solutions in $\bg_white z$ will have modulus of 1 and satisfy $\bg_white |z| = 1$. This is not true for the two real solutions so you run into a problem.