291. Find the set of all points on the $xy$-plane whose coordinates satisfy the following condition: the number $z^2 + z + 1$ is real and positive.

Solution: Let $z = x + iy$ then the equation becomes $x^2 - y^2 + x + 1 + iy(1 + 2x) = 0$.

Clearly, imaginary part has to be zero i.e. either $y = 0$ or $x = -\frac{1}{2}$. So, it is real and positive for all points on the x-axis. When, $x = -\frac{1}{2}$ the real part becomes $y^2 = \frac{3}{4}$. Thus, for points $x = -\frac{1}{2}$ and $-\frac{\sqrt{3}}{2}<y<\frac{\sqrt{3}}{2}$ the required condition is satisfied.

292. Find the real values of the parameter $a$ for which at least one complex number $z$ satisfies the equality $|z - az| = a + 4$ and the inequality $|z - 1|< 1$.

Solution: First equation represents a circle whose center is at $(0, ia)$ and radius equal to $\sqrt{a + 4}$. The second equation represents interior of a circle with center at $(2, 0)$ and radius unity. Now, for the possibility of existence of $z$ the two circles must intersect each other.

$\Rightarrow \sqrt{a^2 + 4} \leq a + 4 + 1 \Rightarrow a \geq -\frac{21}{10}$ and $a + 4 - 1\leq \sqrt{a^2 + 4} \Rightarrow a\leq -\frac{5}{6}$. Combining these two gives us the range for values of $a$.

293. Find the real values of the parameter $t$ for whihc at least one complex number $z$ satisfied the equality $|z + \sqrt{2}| = t^2 - 3t + 2$ and the inequality $|z + i\sqrt{2}| < t^2$.

Solution: Let $z = x + iy$ then $|z + \sqrt{2}| = \sqrt{x^2 + 2\sqrt{2}x + 2 + y^2} = t^2 - 3t + 2$ and $|z + i\sqrt{2}| = \sqrt{x^2 + y^2 + 2\sqrt{2}y + 2} < t^2$.

Because $|z + \sqrt{2}| > 0 \Rightarrow t^2 - 3t + 2 > 0 \Rightarrow t < 1, t> 2$ and $t > 0$. Both the equations are circles so they must intersect for $t$ to exist. The distance between centers i.e. $(-\sqrt{2}, 0)$ and $(0, -i\sqrt{2})$ is $2$.

$\Rightarrow r_1 + r_2 > 2 \Rightarrow 2t^2 - 3t + 2 > 2 \Rightarrow t(2t - 3) > 0 \Rightarrow t < 0, t> \frac{3}{2}$ and $r_1 < r_2 + 2 \Rightarrow t^2 - 2t + 2 < t^2 + 2 \Rightarrow t> 0$. Combining all the inequalities, $t > 2$.

294. Find the real value of $a$ for which there is at least one complex number satisfying $|z + 4i| = \sqrt{a^2 - 12a + 28}$ and $|z - 4\sqrt{3}| < 1$.

Solution: Let $z = x + iy$ then $\sqrt{x^2 + 8x + 16 + y^2} = \sqrt{a^2 - 12a + 28}$ and $\sqrt{x^2 - 8\sqrt{3}x + 48 + y^2} < 1$.

Becaise $|z + 4| > 0 \Rightarrow a^2 - 12a + 28 > 0 \Rightarrow a > 6 + 2\sqrt{2}, a < 6 - 2\sqrt{2}$ and $a> 0$. Both the equations are circles so they must intersect for $a$ to exist. The distance between centers i.e. $(0, -4i)$ and $(4\sqrt{3}, 0)$ is $8$.

$\Rightarrow r_1 + r_2 > 8 \Rightarrow \sqrt{a^2 - 12a + 28} + a > 8 \Rightarrow a > 9$ and $r_1 < r_2 + 8 \Rightarrow a < -\frac{9}{7}$. Combining all these inequalities we have $a> 9$.

295. Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1 + i)z^2$ is positive.

Solution: Let $z = x + iy \Rightarrow (1 + i)z^2 = (1 + i)(x^2 - y^2 + 2ixy)\Rightarrow \Re[(1 + i)z^2] = x^2 - y^2 - 2xy > 0 \Rightarrow x$ has two limits $y(1 \pm\sqrt{2})$.

296. Solve the equation $2z = |z| + 2i$ in complex numbers.

Solution: Let $z = x + iy$ then $2z = |z| + 2i \Rightarrow 2(x + iy) = \sqrt{x^2 + y^2} + 2iy$. Equating real and imaginary parts, $y = 1, 2x = \sqrt{x^2 + 1}$. Squaring $4x^2 = x^2 + 1 \Rightarrow x = \pm\frac{1}{\sqrt{3}}$.

297. Three points represented by the complex numbers $a, b, c$ lie on a circle with center $O$ and radius $r$. The tangent at $c$ cuts the chord joining the points $a, b$ at $z$. Show that $z = \frac{a^{-1} + b^{-1} - 2c^{-1}}{a^{-1}b^{-1} - c^{-2}}$.

Solution: We have earlier proven that if there are two non-parallel lines cutting a circle at $a, b$ and $c, d$ then their point of intersection is given by $\frac{a^{-1} + b^{-1} - c^{-1} - d^{-1}}{a^{-1}b^{-1}- c^{-1}d^{-1}}$.

Now if $c$ and $d$ coincide then that line will become a tangent. So putting $d = c$ we have $z = \frac{a^{-1} + b^{-1} - 2c^{-1}}{a^{-1}b^{-1} - c^{-2}}$.

298. Show that all roots of the equation $a_1z^3 + a_2z^2 + a_3z + a_4 = 3$, where $|a_i|\leq 1, i = 1, 2, 3, 4$ lie outside the circle with center as origin and radius $\frac{2}{3}$.

Solution: Given $a_1z^3 + a_2z^2 + a_3z + a_4 = 3 \Rightarrow |a_1z^3 + a_2z^2 + a_3z + a_4| = 3 \Rightarrow |a_1z^3| + |a_2z^2| + |a_3z| + |a_4| \geq 3$

$\Rightarrow |a_1||z^3| + |a_2||z^2| + |a_3||z| + |a_4|\geq 3 \Rightarrow |z|^3 + |z|^2 + |z| + 1\geq 3\;[\because |a_i|\leq 1]$

$\Rightarrow 1 + |z| + |z|^2 + |z|^3 + \cdots $ to $\infty > 3 \Rightarrow \frac{1}{1 - |z|} > 3 \Rightarrow |z| > \frac{2}{3}$, which shows that roots lie outside the circle with center origin and radius $\frac{2}{3}$.

299. Given that $\displaystyle\sum_{i = 1}^nb_i = 0$ and $\displaystyle\sum_{i=1}^nb_iz_i = 0$, where $b_i$s are non-zero real numbers, no three of $z_i$'s form a straight line. Prove that $z_i$'s are concyclic if $b_1b_2|z_1 - z_2|^2 = b_3b_4|z_3 - z_4|^2$.

Solution: Given, $b_1z_1 + b_3z_3 = -(b_2z_2 + b_4z_4)$ and $b_1 + b_3 = -(b_2 + b_4)\;\therefore \frac{b_1z_1 + b_3z_3}{b_1 + b_3} = \frac{b_2z_2 + b_4z_4}{b_2 + b_4}$.

This means that the point dividing $AC$ in the ratio $b_3:b_1$ also divides $BC$ in the ratio $b_4:b_2$. Let this point be $O$. Let $b_1b_2|z_1 - z_2|^2 = b_3b_4|z_3 - z_4|^2$

$\Rightarrow b_1b_2(b_3^2 + b_4^2 - 2b_3b_4\cos\alpha) = b_3b_4(b_2^2 + b_1^2 - 2b_1b_2\cos\alpha)$

$\Rightarrow \frac{b_3}{b_4} + \frac{b_4}{b_3} = \frac{b_1}{b_2} + \frac{b_2}{b_1}\Rightarrow \frac{b_3}{b_4} = \frac{b_1}{b_2}$ or $\frac{b_2}{b_1}$

If $\frac{b_3}{b_4} = \frac{b_1}{b_2}$, then $\frac{b_3}{b_1} = \frac{b_4}{b_1} \Rightarrow \frac{AO}{CO} = \frac{BO}{DO}$

$\Rightarrow \triangle AOB\sim\triangle BCO \Rightarrow \angle BAO = \angle CDO \Rightarrow AB\parallel CD$ which is not possible.

If $\frac{b_3}{b_4} = \frac{b_2}{b_1}$ then $\frac{AO}{BO} = \frac{DO}{CO}\Rightarrow \triangle ADO\sim\triangle BCO\Rightarrow \angle DAO = \angle OBC\Rightarrow A, B, C, D$ are concyclic.

300. A cubic equation $f(x) = 0$ has one real root $\alpha$ and two complex roots $\beta \pm i\gamma$. Points $A, B$ and $C$ represent these roots. Show that the roots of the derived equation $f'(x) = 0$ are complex if $A$ falls inside one of the two equilateral triangles described on base $BC$.

Solution: Let $f(x) = k(x - a)(x - \beta - i\gamma)(x - \beta + i\gamma) = k(x - a)[(x - \beta)^2 + \gamma^2]$

$\Rightarrow f'(x) = k[3x^2 - 2(a + 2\beta)x + \beta^2 + \gamma^2 + 2a\beta]$. Discriminant of $f'(x)$ is given by $D = 4[(a + 2\beta)^2 - 3(\beta^2 + \gamma^2 + 2a\beta)] = 4(a^2 + \beta^2 - 3\gamma^2 - 2a\beta)$

$BC = 2|\gamma| \Rightarrow PL = \sqrt{3}|\gamma|$. If $A$ lies inside the equilateral triangle having $BC$ as base, then $|\beta - a|< \sqrt{3}\gamma \Rightarrow (\beta - a)^2 < 3\gamma^2 \Rightarrow a^2 + \beta^2 - 3\gamma^2 - 2a\beta < 0 \Rightarrow D < 0$. Thus roots will be complex numbers.