281. Show that the triangles whose vertices are $z_1, z_2, z_3$ and $z_1', z_2', z_3'$ are similar if $\begin{vmatrix} z_1 & z_1' & 1\\ z_2 & z_2' & 1\\ z_3 & z_3' & 1\end{vmatrix} = 0$.

Solution: $\triangle ABC$ and $\triangle PQR$ will be similar if all their angles are equal and ratios of sides as well.

$arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right) = arg\left(\frac{{z^\prime}_3 - {z^\prime}_1}{{z^\prime}_2 - {z^\prime}_1}\right)$

$\frac{AB}{PQ} = \frac{AC}{PR}$ or $\frac{AC}{AB} = \frac{PR}{PQ}$ or $\frac{z_3 - z_1}{z_2 - z_1} = \frac{{z^\prime}_3 - {z^\prime}_1}{{z^\prime}_2 - {z^\prime}_1}$

Simplifying these two equations gives us our determinant.

282. If $a, b, c$ and $u, v, w$ are the complex numbers representing two triangles such that $c = (1 - r)a + rb$ and $w = (1 - r)u + rv$, where $r$ is a complex number, prove that the two triangles are similar.

Solution: From these two equation we have $r = \frac{c - a}{b - a}$ and $r = \frac{\omega - u}{v -u}$. Equating these two equations and taking modulus and argument, it follows from the previous problem that the two triangles are similar.

283. Find the equation of perpendicular bisector of the line segment joining points $z_1$ and $z_2$.

Solution: We know that points on a perpendicular bisector is equidistant from the two points of the line to which it is perperndicular bisector.

$\Rightarrow |z - z_1| = |z - z_2| \Rightarrow |z - z_1|^2 = |z - z_2|^2 \Rightarrow (z - z_1)(\overline{z} - \overline{z_1}) = (z - z_2)(\overline{z} - \overline{z_2})$, which can be written in the form of $\overline{a}z + a\overline{z} + b = 0$, which is equation of a straight line.

284. Find the equation of a circle having the line segment joining $z_1$ and $z_2$ as diameter.

Solution: Mid-point of such a diameter is $\frac{z_1 + z_2}{2}$. Let $P$ be a point lying on this circle, which, is represented by complex number $z$. Thus, the equation of circle is $\left|z - \frac{z_1 + z_2}{2}\right| = \left|z_1 - \frac{z_1 + z_2}{2}\right|$ or $\left|z - \frac{z_1 + z_2}{2}\right| = \left|z_2 - \frac{z_1 + z_2}{2}\right|$. Square and simplify to arrive at the equation.

285. If $\left|\frac{z - z_1}{z - z_2}\right| = c, c\neq 0,$ then show that locus of $z$ is a circle.

Solution: The equation can be written as $\left|z - z_1\right| = c\left|z - z_2\right|$, which, when substituted with $z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$ gives following

$\left|(x - x_1) + i(y - y_1)\right| = c\left|(x - x_2) + i(y - y_2)\right| \Rightarrow (x - x_1)^2 + (y - y_1)^2 = c^2\{(x - x_2)^2 + (y - y_2)^2\}$, which is equation of a circle.

286. If $|z| = 1,$ find the locus of the point $\frac{2}{z}$.

Solution: Given, $|z| = 1 \Rightarrow 2z\overline{z} = 2 \Rightarrow \frac{2}{z} = 2\overline{z}$ which gives us a circle.

287. If for any two complex numbers $z_1$ and $z_2, |z_1 + z_2| = |z_1| + |z_2|$, prove that $\arg(z_1) - \arg(z_2) = 2n\pi$.

Solution: Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$. Then L.H.S. $= \left|z_1 + z_2\right|$ $\Rightarrow \left|z_1 + z_2\right|^2 = {r^2}_1 + {r^2}_2 + 2r_1r_2\cos(\theta_1 - \theta_2)$.

Similarly, $\left(\left|z_1\right| + \left|z_1\right|\right)^2 = \left({r^2}_1 + {r^2}_2 + 2r_1r_2\right)$.

Thus, $\cos(\theta_1 - \theta_2) = 0$ $\Rightarrow \arg(z_1) - \arg(z_2) = 2n\pi$.

288. Find the complex number $z$, the least in absolute value, which satisfies the condition $|z - 2 + 2i| = 1$.

Solution: The equation $\left|z - 2 + 2i\right| = 1$ represents a circle with center at $(2, -2i)$ with unity radius. Since, the line between $(2, -2i)$ and origin will make an angle of $45^\circ$. Therefore, $P$ is $2 - \frac{1}{\sqrt{2}} + i(\frac{1}{\sqrt{2}} -2)$.

289. Find the point in the first quadrant, on the curve $|z - 5i| = 3$, whose argument is minimum.

Solution: Given equation is a circle with center $(0, 5)$ and radius $3 \therefore OC = 5, CP = 3$.

The point having least argument will have a tangent from origin which makes $\triangle OCP$ right angle triangle.

$\Rightarrow CP = 4\Rightarrow \tan\theta = \frac{4}{3}$. Therefore, the point would be $4(\cos\theta + i\sin\theta) = \frac{12}{5} + \frac{16i}{5}$.

290. Find the set of points of the cooradinate plane, which satisfy the inequality $\log_{1/2}\left(\frac{|z - 1| + 4}{3|z - 1| - 2}\right) > 1$

Solution: From given equation, $\left(\frac{\left|z-1\right| + 4}{3\left|z - 1\right| - 2}\right) < \frac{1}{2}$

$\Rightarrow |z - 1| > 10$. This represents area which lies outside a circle with center at $(1, 0)$ and radius $10$.