complex numberslocus
I have solved most of the question. This is my work but I can't seem to find what this part is. It's supposed to be $1\pm i\sqrt{15}$ but I don't know why. Any help is much appreciated. Thank you
On a single Argand diagram, sketch the loci $\lvert z\rvert=4$ and $\lvert z+2\rvert=\lvert z-4\rvert$. Hence determine complex numbers that satisfy both loci, giving your answer in Cartesian form.
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.
Here is a diagram of the problem:
We see that the line containing the origin and the point of least argument $p$ in $B$ forms a tangent to the circle $|z-u|=1$. The segment of this line between the two points has length $2\sqrt2$ and is the hypotenuse of a right triangle with one leg of length 1 (corresponding to the circle's radius). Therefore the length of the other leg, which is also the modulus of $p$ and the answer to the question, is $$\sqrt{(2\sqrt2)^2-1^2}=\sqrt7$$
Best Answer
The equation: $|z+2|=|z-4|$ is the locus of perpendicular bisector of line segment joining $(-2,0)$ and $(4,0)$. So it is straight line whose equation is: $x=(4-2)/2=1$.
The equation $|z|=4$ represents a circle of radius 4 centered at $(0,0)$. Put $z=x+iy=1+iy$ in the circle’s equation to get point of intersection.