complex numbersfoundations
Find all complex numbers $z$ such that $|z|=|\sqrt{2} +z|=1$. Prove that each of these satisfies $z^8=1$ I'm a little unsure of where to start. I know that the modulus of $z$ is the distance from the origin. The book states
$|z|=1$ implies $z=\cos\theta + i\sin\theta$ for some $\theta$
This part makes sense. But the part I don't understand is this:
$|\sqrt{2} +z|=1$ implies $(\cos\theta + \sqrt{2})^2 +\sin^2\theta = 1$
Any help would be appreciated.
In general, the angle $\theta$ between two complex vectors $u$ and $v$ (itself possibly complex) is given by $$ \theta = \cos^{-1}\frac{u\cdot{v^H}}{\left\lVert{u}\right\rVert\left\lVert{v}\right\rVert} $$ For the example above, $v^H$ = $(2+3i,4-i)$ and $u\cdot{v^H}$ = $6 - i$. The norm of a complex vector $u = \{u_1, u_2, \dots, u_n\}$ is given by $$ \left\lVert{u}\right\rVert = \sqrt{\sum_{i=1}^n\ u_iu_i^*} = \sqrt{\sum_{i=1}^n\ \left|u_i\right|^2} $$ This, of course, reduces to the standard (Euclidean) norm for real vectors. We see for $u$ and $v$ above the norms are $\left\lVert{u}\right\rVert = \sqrt{7}$ and $\left\lVert{v}\right\rVert = \sqrt{30}$ respectively, yielding a value for the angle between the vectors of $$ \theta = \cos^{-1}\frac{6-i}{\sqrt{210}} \approx 1.14521 + 0.0756928i $$
The diagram is on the right track, but you will also have to do a translation given the fact that $y=x-2$.
The steps as I would envision are:
Make the following substitutions: $$ x = \frac{z+ \overline{z}}{2}$$ $$ y = \frac{z- \overline{z}}{2i}$$
Apply the rotational coordinate transformation: $$ z' = ze^{-i\theta}$$ $$\theta = artcan(m)$$
Apply the translation coordinate transformation: $$ z'' = z' + ai $$ $$ a = \sqrt{2} $$
Perform the reflection: $$ z'' = \overline{z''}$$
'Undo' the translation coordinate transformation: $$ z' = z'' - ai $$
'Undo' the rotational coordinate transformation: $$ z = z'e^{i\theta}$$
Diagram of translation first approach (Ellipse position is very approximate):
Best Answer
The intersection occurs at $z=x\pm iy$, where $x=-\frac12\sqrt2$ and $x^2+y^2=1\implies y=\frac12\sqrt2$