# Thread: Roots of an Equation

1. ## Roots of an Equation

This is my first time here, so I'm not sure if this question belongs here or somewhere else. I'm actually in Year 11, but none of the pre-university topics seemed to cover this, so sorry if I posted this in the wrong place:

Show that the roots of (z-1)^4 + (z+1)^4 = 0 are +/- i cot (pi/8) and +/- i cot (3pi/8)

Thanks

2. Originally Posted by st123
This is my first time here, so I'm not sure if this question belongs here or somewhere else. I'm actually in Year 11, but none of the pre-university topics seemed to cover this, so sorry if I posted this in the wrong place:

Show that the roots of (z-1)^4 + (z+1)^4 = 0 are +/- i cot (pi/8) and +/- i cot (3pi/8)

Thanks
As $z=-1$ is NOT a solution, we can do:

$(z-1)^4+(z+1)^4=0\Longleftrightarrow \frac{(z-1)^4}{(z+1)^4}=-1\Longleftrightarrow \left(1 - \frac{2}{z+1}\right)^4=-1$.

Solve now $w^4=-1=e^{\pi i+2k\pi i}=cis(\pi+2k\pi)\,,\,k=0,1,2,3\Longrightarrow w=e^{\frac{\pi i}{4}\left(2k+1\right)}$ , using De Moivre's Theorem, and once you get the roots

just substitute $w = 1-\frac{2}{z+1}$ and solve for $z$ .

If you don't understand something of the above then this question isn't for you, yet. Complex numbers enter usually in 11th or 12th years (5ht or 6th high school).

Tonio

3. Originally Posted by st123
This is my first time here, so I'm not sure if this question belongs here or somewhere else. I'm actually in Year 11, but none of the pre-university topics seemed to cover this, so sorry if I posted this in the wrong place:

Show that the roots of (z-1)^4 + (z+1)^4 = 0 are +/- i cot (pi/8) and +/- i cot (3pi/8)

Thanks
Of course you could just try substituting the purported roots into the left hand side and showing that the result is zero.

CB

4. $(z - 1)^4 + (z + 1)^4 = z^4 - 4z^3 + 6z^2 - 4z + 1 + z^4 + 4z^3 + 6z^2 + 4z + 1$

$= 2z^4 + 12z^2 + 2$

$= 2(z^4 + 6z^2 + 1)$

$= 2\left(z^4 + 6z^2 + 3^2 - 3^2 + 1\right)$

$= 2\left[(z^2 + 3)^2 - 8\right]$

Therefore, if $(z - 1)^4 + (z + 1)^4 = 0$

then $2\left[(z^2 + 3)^2 - 8\right] = 0$

$(z^2 + 3)^2 - 8 = 0$

$(z^2 + 3)^2 = 8$

$z^2 + 3 = \pm 2\sqrt{2}$

$z^2 = -3 \pm 2\sqrt{2}$

$z = \pm \sqrt{-3 \pm 2\sqrt{2}}$.

Now, you should note that since $2\sqrt{2} < 3$, that means $-3 + 2\sqrt{2} < 0$ and $-3 - 2\sqrt{2} < 0$.

Therefore $z = \pm \sqrt{-3 \pm 2\sqrt{2}}$

$z = \pm\sqrt{-(3 \mp 2\sqrt{2})}$

$z = \pm i\sqrt{3 \pm 2\sqrt{2}}$.

You should be able to show that these values are the same as what you have written.

5. Originally Posted by tonio
As $z=-1$ is NOT a solution, we can do:

$(z-1)^4+(z+1)^4=0\Longleftrightarrow \frac{(z-1)^4}{(z+1)^4}=-1\Longleftrightarrow \left(1 - \frac{2}{z+1}\right)^4=-1$.

Solve now $w^4=-1=e^{\pi i+2k\pi i}=cis(\pi+2k\pi)\,,\,k=0,1,2,3\Longrightarrow w=e^{\frac{\pi i}{4}\left(2k+1\right)}$ , using De Moivre's Theorem, and once you get the roots

just substitute $w = 1-\frac{2}{z+1}$ and solve for $z$ .

If you don't understand something of the above then this question isn't for you, yet. Complex numbers enter usually in 11th or 12th years (5ht or 6th high school).

Tonio
Thanks for helping me out, your solution was the most 'elegant'.