251. If $\alpha, \beta, \gamma, \delta$ be the real roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$, show that $a^2(1 + \alpha^2)(1 + \beta^2)(1 + \gamma^2)(1 + \delta^2) = (a - c + e)^2 + (b - d)^2$.

Solution: Since $\alpha, \beta, \gamma, \delta$ are roots of the equation. $\therefore a(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = ax^4 + bx^3 + cx^2 + dx + e$.

Substituting $x = i$, we get following

$a(i - \alpha)(i - \beta)(i - \gamma)(i - \delta) = ai^4 + bi^3 + ci^2 + di + e \Rightarrow a(1 + i\alpha)(1 + i\beta)(1 + i\gamma)(1 + i\delta) = a - ib - c + id + e$.

Taking modulus and squaring we get our desired result.

252. If $a_i\in R, i = 1, 2, \ldots, n$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ are the roots of the equation $x^n + a_1x^{n - 1} + a_2x^{n - 2} + \ldots + a_{n- 1}x + a_n = 0$, show that $\displaystyle\prod_{i = 1}^n(1 + \alpha_i^2) = (1 - a_2 + a_4 - \ldots)^2 + (a_1 - a_3 + \ldots)^2$.

Solution: $\because \alpha_1, \alpha_2, \ldots, \alpha_n$ are the roots of the given equation.

$\therefore (x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_n) = x^n + a_1x^{n - 1} + a_2x^{n - 2} + \ldots + a_{n- 1}x + a_n= 0$.

Substituting $x = i$, we get following $(i - \alpha_1)(i - \alpha_2)\cdots(i - \alpha_n) = i^n + a_1i^{n - 1} + a_2i^{n - 2} + \ldots + a_{n- 1}i +a_n$.

Taking modulus and squaring we get our desired result.

253. If the complex numbers $z_1, z_2, z_3$ are the vertices of an equilateral triangle such that $|z_1| = |z_2| = |z_3|$, prove that $z_1 + z_2 + z_3 = 0$.

Solution: Let $|z_1| = |z_2| = |z_3| = R. \therefore$ Origin is the circumcenter of triangle.

Since triangle is also equilateral circumcenter and origin coincide. Therefore, origin is also centroid. Thus,

$\frac{z_1 + z_2 + z_3}{3} = 0 \Rightarrow z_1 + z_2 + z_3 = 0$.

254. If $z_1 + z_2 + z_3 = 0$ and $|z_1| = |z_2| = |z_3| = 1$, then prove that the complex numbers $z_1, z_2, z_3$ are the vertices of an equilateral triangle inscribed in a unit circle.

Solution: $z_1 + z_2 + z_3 = 0$ implies centroid of the triangle is the origin. Circumcenter is also origin as $z_i$ lies on the circle $|z| = 1$.

Hence, circumcenter is same as centroid making the triangle an equilateral triangle having circumcircle with unit radius.

255. If $z_1, z_2, z_3$ be the vertices of an equilateral triangle whose circumcenter is $z_0$, then prove that $z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.

Solution: Since the triangle is equilateral therefore the circumcenter and centroid will be same i.e. $z_0 = \frac{z_1 + z_2 + z_3}{3}$. Also for equilateral triangle, $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$.

Squaring the first equation $9z_0^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_2) = z_1^2 + z_2^2 + z_3^2 + 2(z_1^2 + z_2^2 + z_3^2) \Rightarrow z_1^2 + z_2^2 + z_3^2 = 3z_0^2$.

256. Prove that the complex numbers $z_1$ and $z_2$ and the origin form an equilateral triangle if $z_1^2 + z_2^2 - z_1z_2 = 0$.

Solution: Since $z_1, z_2$ and origin form an equilateral triangle we have $z_1^2 + z_2^2 + 0^2 - z_1z_2 - z_2.0 - z_1.0 = 0$. Hence, proven.

257. If $z_1$ and $z_2$ be the roots of the equation $z^2 + az + b = 0$, then prove that the origin, $z_1$ and $z_2$ form an equilateral triangle if $a^2 = 3b$.

Solution: From previous probelm $z_1, z_2$ and origin will form a triangle if $z_1^2 + z_2^2 - z_1z_2 = 0.$ Therefore, $(z_1 + z_2)^2 = 3z_1z_2 \Rightarrow a^2 = 3b$.

258. Let $z_1, z_2$ and $z_3$ be the roots of the equation $z^3 + 3\alpha z^2 + 3\beta z + \gamma = 0$, where $\alpha, \beta$ and $\gamma$ are complex numbers and that these represent the vertices of $A, B$ and $C$ of a triangle. Find the centroid of $\triangle ABC$. Show that the triangle will be equilateral, if $\alpha^2 = \beta$.

Solution: Since $z_1, z_2, z_3$ are roots of the equation $z^3 + 3\alpha z^2 + 3\beta z + \gamma = 0\Rightarrow z_1 + z_2 + z_3 = -3\alpha, z_1z_2 + z_2z_3 + z_3z_1 = 3\beta$ and $z_1z_2z_3 = -\gamma$.

Centroid is given by $\frac{z_1 + z_2 + z_3}{3} = -\alpha$. Triangle will be equilateral if $z_1^2 + z_2^2 + z_3^3 = z_1z_2 + z_2z_3 + z_3z_1 \Rightarrow (z_1 + z_2 + z_3)^2 = 3(z_1z_2 + z_2z_3 + z_3z_1) \Rightarrow \alpha^2 = \beta$.

259. If $z_1, z_2, z_3$ are in A.P., prove that they are collinear.

Solution: Given $2z_2 = z_1 + z_3$. Clearly, from section formula we can deduce that $z_2$ divides line segment joining $z_1$ and $z_3$ in two equal segments hence the complex numbers are collinear.

260. If $z_1, z_2$ and $z_3$ are collinear points in argand plane then show that one of the following holds: $-z_1|z_2 - z_3| + z_2|z_3 - z_1| + z_3|z_1 - z_2| = 0, z_1|z_2 - z_3| - z_2|z_3 - z_1| + z_3|z_1 - z_2| = 0, z_1|z_2 - z_3| + z_2|z_3 - z_1| - z_3|z_1 - z_2| = 0$.

Solution: If $z_1, z_2, z_3$ are collinear then either $z_2$ divides $z_1z_3$ internally/externally or $z_3$ divides $z_1z_2$ internally/externally. Now we can apply the condition for collinearity i.e. $\begin{vmatrix} z_1& z_2 & z_3\\ \overline{z_1} & \overline{z_2} & \overline{z_3}\\ 1& 1& 1\end{vmatrix} = 0$ and hence we can show desired conditions.