1. Complex Locus

z = x + iy, which satisfies

z^2 + z^ -2 = 8

Find the locus of P in terms of x and y.

Is the general aproach to get a equantion in x and y?

If so, i tried this and ive got z^4 - 8z^2 + 1 = 0
Do i just sub in x + iy as z and expand to get locous?

If not what do i do?

2. Originally Posted by Lukybear
z = x + iy, which satisfies

z^2 + z^ -2 = 8

Find the locus of P in terms of x and y.

Is the general aproach to get a equantion in x and y?

If so, i tried this and ive got z^4 - 8z^2 + 1 = 0
Do i just sub in x + iy as z and expand to get locous?
You could do that, but it's likely to be rather messy.
If not what do i do?
I would suggest substituting $w := z^2$ first. If you do that you'll find that $w_{1,2}=4\pm \sqrt{15}$ are the only solutions. Now solve the two equations $(x+iy)^2=4\pm \sqrt{15}$ and you are done.

3. Hello, Lukybear!

$z \,=\, x + iy\text{, which satisfies: }\;z^2 + z^{-2} \:=\: 8$

$\text{Find the locus of }P\text{ in terms of }x\text{ and }y.$

The problem is not clearly stated.
I assume that we want the locus of $z.$

We have: . $z^4 - 8z^2 + 1 \:=\:0$

Quadratic Formula: . $z^2 \:=\:\dfrac{8\pm\sqrt{60}}{2} \:=\:\dfrac{8\pm2\sqrt{15}}{2} \:=\:4 \pm\sqrt{15}$ .[1]

We note that: . $(\sqrt{5}\pm \sqrt{3})^2 \;=\;5 \pm 2\sqrt{15} + 3 \;=\;8 \pm 2\sqrt{15} \;=\;2(4 \pm \sqrt{15})$

. . Hence: . $4 \pm \sqrt{15} \:=\:\dfrac{(\sqrt{5} \pm \sqrt{3})^2}{2}$

Then [1] becomes: . $z^2 \;=\;\dfrac{(\sqrt{5}+\sqrt{3})^2}{2}$

. . Hence: . $z \;=\;\pm\left(\dfrac{\sqrt{5}\pm\sqrt{3}}{\sqrt{2} }\right)$ . . . all real numbers.

The locus is four vertical lines:

. . $\begin{array}{ccccc}
x &=& \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}} & \approx & 2.806\\ \\[-3mm]
x &=& \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & 0.356\\ \\[-3mm]
x &=& \dfrac{-\sqrt{5} + \sqrt{3}}{\sqrt{2}} & \approx & \text{-}0.356\\ \\[-3mm]
x &=& \dfrac{-\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & \text{-}2.806 \end{array}$

4. Ok thanks. Much easier than i had anticpiated.

5. Originally Posted by Soroban
The locus is four vertical lines:

. . $\begin{array}{ccccc}
x &=& \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}} & \approx & 2.806\\ \\[-3mm]
x &=& \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & 0.356\\ \\[-3mm]
x &=& \dfrac{-\sqrt{5} + \sqrt{3}}{\sqrt{2}} & \approx & \text{-}0.356\\ \\[-3mm]
x &=& \dfrac{-\sqrt{5} - \sqrt{3}}{\sqrt{2}} & \approx & \text{-}2.806 \end{array}$
No, I don't believe that that's quite right. Going back to $z^2=4\pm\sqrt{15}$ and setting $z := x+iy$, we get $(x+iy)^2=x^2+2ixy-y^2=4\pm\sqrt{15}$.
Comparing real- and imaginary parts of the left and right hand side we see that we must have $xy=0$ and $x^2-y^2=4\pm\sqrt{15}$.

But $x=0$ is not possible, because then the second of these two equations would not have a (real) solution y.

Setting $y=0$ to satisfy the first equation gives the solutions $x_{1,2,3,4}=\pm\sqrt{4\pm\sqrt{15}}$ of the second equation.

Hence I get $z_{1,2,3,4}=\pm\sqrt{4\pm\sqrt{15}}$ four isolated points (and not four vertical lines).

6. So what is the locus?

7. Originally Posted by Lukybear
So what is the locus?
As I wrote, it is just the four points $z_{1,2,3,4}=\pm\sqrt{4\pm \sqrt{15}}$ on the real axis. This agrees with the solution Soroban gave you, with the small difference that y cannot be arbitrary, but must be =0.

If you take $z:= \sqrt{4+\sqrt{15}}+i\cdot 2$, for example, which would lie on one of those vertical axes Soroban found, you get that the left hand side of the original equation evaluates to $z^2+1/z^2\approx 3.90+i\cdot 11.14$, which is rather too far away from the value of the right hand side, $8$.