complex numbers
Sorry I am back again so soon but I am struggling to understand the following question:
Find three complex numbers that satisfy the following equation: |z-2| = |z-3i|
So far I believe I have two correct solutions which are $z = \frac{-5}{4}$ and $1 +\frac{3i}{2}$
but I am at a loss on how to find the third solution.
Thank you.
I suppose the most common one on this site is an application of the Residue Theorem. That is:
$$\int_\gamma f(z) dz = 2\pi i \sum_k Res(f; z_k)$$
where $f$ is an analytic function with only finitely many isolated singularities $z_k$ inside a closed curve $\gamma$ in the complex plane.
While this theorem is clearly a result of Complex Analysis, it in fact has many uses in computing integrals along the real line. Indeed, by constructing $\gamma$ to be semicircular contours, we can immediately compute the real integral $\int_{-\infty}^\infty f(x) dx$ for functions $f(z)$ that are the complex extension of real-valued $f(x)$ (as long as $f(z)$ disappears as $|z|\rightarrow \infty$).
This typical contour $\gamma$ appears as:
where $j$ is an isolated singularity of $f(z)$ and we take $a\rightarrow \infty$.
Here is a straight-forward example. We attempt to compute the definite integral:
$$\int_{-\infty}^\infty \cfrac{dx}{(1+x^2)^2}$$
Defining $f(z):= \cfrac{1}{(1+z^2)^2} = \cfrac{1}{(z+i)^2(z-i)^2}$ where $z\in \mathbb{C}$, and the complex contour $\gamma_a$ to be the semicircle in the upper-half plane, we have by the Residue Theorem: $$\int_{\gamma_a} f(z) dz = 2\pi i Res(f; i) = \cfrac{2\pi i}{4i} = \cfrac{\pi}{2}$$
Now, noting that as $|z|\rightarrow \infty, |f(z)| \rightarrow 0$, so
$$ \cfrac{\pi}{2} = \lim_{a\rightarrow \infty} \int_{\gamma_a} f(z) dz = \lim_{a\rightarrow \infty} \left(\int_{-a}^a f(x) dx + \int_{|z|=a,\theta \in [0,\pi]} f(z) dz \right) = \int_{-\infty}^\infty f(x) dx$$
and we have computed our real-valued integral of a real-valued function using Complex Analysis.
$z^3+8=0;$
$(z+2)(z^2-2z +2^2)=0;$
$z_1=-2;$
Solve quadratic equation:
$z_{2,3} = \dfrac{2\pm \sqrt{4-(4)2^2}}{2}$;
$z_{2,3}= \dfrac{2\pm i 2√3}{2}.$
Best Answer
$z=x+iy\implies$
$$(x-2)^2+y^2=x^2+(y-3)^2\iff4x-6y+5=0$$
So, any point of the straight line $4x-6y+5=0$ will satisfy this.
Choose arbitrary $x$ to find one corresponding $y$ OR vice versa.