complex numbers
Sketch, on a single Argand diagram, the loci given by
(i) $| z − \sqrt 3 − i | = 2$
(ii) $\arg (z) = 16 \pi$
Does anyone know a good way of how to explain solving a problem similar or the same as this?
The part $| z − \sqrt3 − i | = 2$ is just a circle at $(\sqrt3,1)$ that crosses the point $(0,0)$ [why?] and is its radius $\sqrt2$?
Second part I don't understand at all.
Well... you have that $x-2 = -y$, where $y>0$. This is just a line. Well, $w \equiv (-v+2) + vi$ and $z = \cos(\theta) + (\sin(\theta)-1)i$. So, $|z-w|^2 = (-v+2-\cos(\theta))^2 + (v+1-\sin(\theta))^2$, so we can easily optimise by choosing $v-\sin(\theta) = 1$ and $v+\cos(\theta) = 2$ yields $\cos(\theta) + \sin(\theta) = 1$ and so $\sin(\theta+\pi/4) = 1/\sqrt{2}$ and so $\theta = 0$ or $\theta = \pi/2$. This then implies that $v = 1$ or $v = 2$ respectively. Now fill in the details.
$\qquad\qquad\qquad$
See the above diagram. Here, note that $\triangle{ABC},\triangle{AOD}$ and $\triangle{DEF}$ are triangles with $45^\circ,45^\circ,90^\circ$ and with $|AC|=|AO|=|OD|=1,|ED|=|EO|-|OD|=2-1=1$.
Thus, $$\begin{align}\text{the least value of $|z-w|$}&=|CF|\\&=|CD|+|DF|\\&=|AD|-|AC|+\frac{1}{\sqrt 2}|DE|\\&=\color{red}{\frac{3}{2}\sqrt 2-1}\end{align}$$
Best Answer
For your first part, let $z=x+iy$ so then we get that $$ | z- (\sqrt{3} + i)| = | (x-\sqrt{3}) + i(y-1)| = (x-\sqrt{3})^2 + (y-1)^2 = 2 $$
which is the locus of a circle with centre $(\sqrt{3},1)$ and radius $ \sqrt{2}$. Substituting $ x= 0, y=0$ gives us that it intersects the origin.
For your second part, consider that arg z = $arctan(\frac{y}{x})$.