261. What region in the argand plane is represented by the inequality $1 < |z - 3 - 4i| < 2$.

Solution: $z$ represents the ring between the concentric circles whose center is at $(3, 4i)$ having radii $1$ and $2$.

262. Find the locus of point $z$ if $|z - 1| + |z + 1|\leq 4$.

Solution: Let $z = x + iy \Rightarrow |z|^2 = x^2 + y^2, |z - 1|^2 = (x - 1)^2 + y^2, |z + 1|^2 = (x + 1)^2 + y^2$.

From given inequailty $|z + 1|^2 = 16 + |z - 1|^2 - 8|z - 1|\Rightarrow 4x = 16 - 8|z - 1|\Rightarrow 4|z - 1|^2 = (4 - x)^2 \Rightarrow 3x^2 + 4y^2 = 12$, which is equation of an ellipse.

263. If $z = t + 5 + i\sqrt{4 - t^2}$ and $t$ is real, find the locus of $z$.

Solution: Let $z = x + iy$, then $x = t + 5 \Rightarrow x - 5 = t$ and $y = \sqrt{4 - t^2}\Rightarrow y^2 = 4 - t^2 \Rightarrow (x - 5)^2 + y^2 = 4$, which is a circle with center $(5, 0)$ and radius $2$.

264. If $\frac{z^2}{z - 1}$ is real, show that locus of $z$ is a circle with center $(1, 0)$ and radius unity.

Solution: Let $z = x + iy$, then $\frac{z^2}{z - 1} = \frac{(x^2 - y^2 + 2ixy)[(x - 1) - iy]}{(x - 1)^2 + y^2}$. Since it is real, we can equate the imaginary part to zero.

$\Rightarrow y(y^2 - x^2) + 2x^2y - 2xy = 0 \Rightarrow y = 0$ or $x^2 + y^2 - 2x = 0 \Rightarrow (x - 1)^2 + y^2 = 1$.

However, $y\neq 0$ else $z$ won't remain a complex number. $\Rightarrow (x - 1)^2 + y^2 = 1$, which represents a circle with center at $(1, 0)$ and unit radius.

265. If $|z^2 - 1| = |z|^2 + 1$, show that locus of $z$ is a straight line.

Solution: Let $z = x + iy$, then $|z^2 - 1| = |z|^2 + 1 \Rightarrow (x^2 - y^2 - 1)^ + 4x^2y^2 = (x^2 + y^2 + 1)^2 \Rightarrow x = 0$.

Hence, locus of $z$ is a straight line specifically imginary axis.

266. Find the locus of the point $z$ if $\frac{\pi}{3}\leq \arg(z)\leq \frac{3\pi}{2}$.

Solution: Let $z = x + iy$ then $\frac{y}{x} \geq \tan\frac{\pi}{3} \Rightarrow y\geq \sqrt{3}x$. Similarly, $\frac{y}{x} \leq \tan\frac{3\pi}{2}=-\infty$.

This represents the set of straight lines whose slope is greater than $\sqrt{3}$ and less than or equal to $-\infty$.

267. Find the locus of the point $z$ if $\arg\left(\frac{z - 2}{z + 2}\right) = \frac{\pi}{3}$.

Solution: Let $z = x + iy$, then $\arg\left(\frac{z - 2}{z + 2}\right) = \frac{\pi}{3} \Rightarrow \arg\left(\frac{x - 2 + iy}{x + 2 + iy}\right) = \frac{\pi}{3}$

$\Rightarrow \arg\left(\frac{x^2 + y^2 - 4 + 4iy}{(x + 2)^2 + y^2}\right) = \frac{\pi}{3}\Rightarrow \frac{4y}{x^2 + y^2 - 4} = \sqrt{3}$, which is equation of a circle.

268. Show that the locus of the point $z$ satisfying the condition $\arg\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{2}$ is the semicircle above $x$-axis, whose diameter is the joints of the points $(-1, 0)$ and $(1, 0)$ excluding these points.

Solution: Let $z = x + iy$. Given, $\arg\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{2} \Rightarrow \arg\left(\frac{(x - 1) + iy}{(x + 1) + iy}\right) = \frac{\pi}{2} \Rightarrow \frac{2y}{x^2 + y^2 - 1} = \infty$.

The above equation implies $x^2 + y^2 - 1 = 0$ and $y > 0$ which is circle at $(0, 0)$ with unit circle above $x$-axis. The points $(-1, 0)$ and $(1, 0)$ are excluded because that will make the above equation indeterminate.

269. Find the locus of the point $z$ if $\log_{\sqrt{3}}\frac{|z|^2 - |z| + 1}{2 + |z|} < 2$.

Solution: $\log_{\sqrt{3}}\frac{|z|^2 - |z| + 1}{2 + |z|} < 2\Rightarrow \frac{|z|^2 - |z| + 1}{2 + |z|} < (\sqrt{3})^2\Rightarrow |z|^2 - 4|z| - 5 < 0\Rightarrow 0 < |z| < 5$, which is interior of the circle having radius of $5$.

270. If $O$ be the center of the circle circumscribing the equilateral $\triangle ABC$ and its radius be unity and $A$ lies on the $x$-axis. Find the complex numbers represented by $B$ and $C$.

Solution: Clearly $A$ is $(1, 0)$ or $(-1, 0)$. Let A is (1, 0). Then $z = \cos0^\circ + i\sin0^\circ.$ Clearly, $B$ and $C$ would be $\cos120^\circ + i\sin120^\circ$ and $\cos240^\circ + i\sin240^\circ$. Similarly, $B$ and $C$ can be found if $A$ is $(-1, 0)$.