301. Prove that the reflection of $\overline{a}z + a\overline{z} = 0$ in the real axis is $\overline{az} + az = 0$.

Solution: Let $a = \alpha + i\beta$ and $z = x + iy$, then $\overline{a}z + a\overline{z} = 0$ becomes as $\alpha x + \beta y = 0$ or $y = \left(\frac{-\alpha}{\beta}\right)x$.

Its reflection in the x-axis is $y = \frac{\alpha}{\beta}x$ or $\alpha x - \beta y = 0 \Rightarrow \left(\frac{a + \overline{a}}{2}\right)\left(\frac{z + \overline{z}}{2}\right) - \left(\frac{a - \overline{a}}{2}\right)\left(\frac{z -\overline{z}}{2}\right) = 0$

$\Rightarrow az + \overline{a}\overline{z} = 0$

302. If $\alpha, \beta, \gamma, \delta$ are four complex numbers such that $\frac{\gamma}{\delta}$ is real and $\alpha\delta - \beta\gamma \neq 0$, then prove that $z = \frac{\alpha + \beta t}{\gamma + \delta t}, t\in R$ represents a straight line.

Solution: We have $z = \frac{\alpha + \beta t}{\gamma + \delta t} \Rightarrow t = \frac{\alpha - \gamma z}{\delta z - \beta}$. As $t$ is real, $\frac{\alpha - \gamma z}{\delta z - \beta} = \frac{\overline{\alpha} - \overline{\gamma z}}{\overline{\delta z} - \overline{\beta}}$

$\Rightarrow \Rightarrow (\alpha - \gamma z)(\overline{\delta z} - \overline{\beta}) = (\overline{\alpha} - \overline{\gamma z})(\delta z - \beta)$

$\Rightarrow (\overline{\gamma}\delta - \gamma\overline{\delta})z\overline{z} + (\gamma\overline{\beta}-\overline{\alpha}\delta)z +(\alpha\overline{\delta} - \beta\overline{\gamma})\overline{z} = (\alpha\overline{\beta} - \overline{\alpha}\beta)$

Since $\frac{\gamma}{\delta}$ is real, $\frac{\gamma}{\delta} = \frac{\overline{\gamma}}{\overline{\delta}}$ or $\gamma\overline{\delta} - \delta\overline{\gamma} = 0$.

Thus, $\overline{a}z + a\overline{z} = c$, where $a = i(\alpha\overline{\delta}) - \beta\overline{\gamma}$ and $c = i(\overline{\alpha}\beta - \alpha\overline{\beta})$.

Note that $a \ne 0$ for if $a = 0$ then $\alpha\overline{\delta} - \beta\overline{\gamma} = 0\Rightarrow \frac{\alpha}{\beta} = \frac{\overline{\gamma}}{\overline{\delta}} = \frac{\gamma}{\delta}\Rightarrow \alpha\delta - \beta\gamma = 0$, which is against the hypothesis.

Also, note that $c = i(\overline{\alpha}\beta - \alpha\overline{\beta})$ is a purely real number. Thus, $z = \frac{\alpha + \beta t}{\gamma + \delta t}$ represents a straight line.

303. If $\omega, \omega^2$ are cube roots of unity, then prove that

  1. $(3 + 3\omega + 5\omega^2)^6 - (2 + 6\omega + 2\omega^2)^3 = 0$.
  2. $(2 - \omega)(2 - \omega^2)(2 - \omega^{10})(2 - \omega^{11}) = 49$.
  3. $(1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^5) = 9$.
  4. $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 = 32$.
  5. $1 + \omega^n + \omega^{2n} = 3$, where $n >0, n\in I$ and is a multiple of $3$.
  6. $1 + \omega^n + \omega^{2n} = 0$, where $n >0, n\in I$ and is not a multiple of $3$.

Solution: The solutions are given below:

  1. L.H.S. $= (3 + 3\omega + 5\omega^2)^6 - (2 + 6\omega + 2\omega^2)^3 = [(3 + 3\omega + 3\omega^2 + 2\omega^2)^6 - (2 + 2\omega + 2\omega^2 + 4\omega)^3] = [\{3(1 + \omega + \omega^2) + 2\omega^2\}^6] - [\{2(1 + \omega + \omega^2) + 4\omega\}^3]$

    $= 64\omega^{12} - 64\omega^3 = 0 =$ R.H.S. $[\because 1 + \omega + \omega^2 = 0]$.

  2. L.H.S. $= (2 - \omega)(2 - \omega^2)(2 - \omega^{10})(2 - \omega^{11}) = (2 - \omega)(2 - \omega^2)(2 - \omega)(2 - \omega^2) = [(2 - \omega)(2 - \omega^2)]^2$

    $= (4 - 2\omega - 2\omega^2 + \omega^3)^2 = [5 - 2(\omega + \omega^2)]^2 = (5 + 2)^2 = 49 =$ R.H.S.

  3. L.H.S. $= (1 - \omega)(1 - \omega^2)(1 - \omega^4)(1 - \omega^5) = (1 - \omega)^2(1 - \omega^2)^2 = (1 - \omega - \omega^2 + \omega^3)^2$ $= [2 - (-1)]^2 = 9 =$ R.H.S.
  4. L.H.S. $= (1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5 = (-2\omega)^5 + (-2\omega^2)^5 = -32(\omega + \omega^2) = 32 =$ R.H.S.
  5. L.H.S. $= 1 + \omega^n + \omega^{2n}$, where $n = 3m\;\forall\;m\in \mathbb{I}$ L.H.S. $= 1 + \omega^{3m} + \omega^{6m} = 1 + (\omega^3)^m + (\omega^3)^{2m} = 1 + 1 + 1 = 3 =$ R.H.S.

  6. We have to prove that $1 + \omega^n + \omega^{2n} = 0$. If $n = 3m + 1\;\forall\;m\in\mathbb{I}$ then L.H.S. $= 1 + \omega^{3m + 1} + \omega^{6m + 2} = 1 + \omega + \omega^2 = 0 =$ R.H.S.

    If $n = 3m + 2,\;\forall\;m\in\mathbb{I}$ then L.H.S. $= 1 + \omega^{3m + 2} + \omega^{6m + 4} = 1 + \omega^2 + \omega = 0 =$ R.H.S.

304. Resolve into linear factors $a^2 + b^2 + c^2 - ab - bc - ca$.

Solution: We have $a^2 + b^2 + c^2 - ab - bc - ca = a^2 + \omega^3b^2 + \omega^3c^2 + (\omega + \omega^2)ab + (\omega + \omega^2)bc + (\omega + \omega^2)ca$

$= (a^2 + ab\omega + ca\omega^2) + (ab\omega^2 + b^2\omega^3 + bc\omega) + (ca\omega + bc\omega^2 + c^2\omega^3)$

$= a(a + b\omega + c\omega^2) + b\omega^2(a + b\omega + c\omega^2) + c\omega(a + b\omega + c\omega^2)$

$= (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.

305. If $x = a + b, y = a\omega + b\omega^2, z = a\omega^2 + b\omega$, prove that $x^3 + y^3 + z^3 = 3(a^3 + b^3)$ and $xyz = a^3 + b^3$.

Solution: $x^3 + y^3 + z^3 = (a + b)^3 + (a\omega + b\omega^2)^3 + (a\omega^2 + b\omega)^3 = a^3 + b^3 + 3a^2b + 3ab^2 + a^3\omega^3 + b^3\omega^6 + 3a^2b\omega^4 + 3ab^2\omega^5 + a^3\omega^6 + b^3\omega^3 + 3a^2b\omega^5 + 3ab^2\omega^4 = 3[a^3 + b^3 + 3a^2b(1 + \omega + \omega^2) + 3ab^2(1 + \omega + \omega^2)] = 3(a^3 + b^3) = $R.H.S.

$xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega) = (a + b)(a^2 + ab\omega + ab\omega^2 + b^2) = (a + b)(a^2 + b^2 - ab) = a^3 + b^3 =$ R.H.S.

306. Resolve into linear factors:

  1. $a^2 - ab + b^2$
  2. $a^2 + ab + b^2$
  3. $a^3 + b^3$
  4. $a^3 - b^3$
  5. $a^3 + b^3 + c^3 - 3abc$

Solution: Given below are the factorization of the expressions:

  1. $a^2 - ab + b^2 = a^2 + (\omega + \omega^2)ab + b^2\omega^3 = (a + b\omega)(a + b\omega^2)$.
  2. $a^2 + ab + b^2 = a^2 - (\omega + \omega^2)ab + b^2\omega^3 = (a - b\omega)(a - b\omega^2)$.
  3. $a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (a + b)(a + b\omega)(a + b\omega^2)$.
  4. $a^3 - b^3 = (a - b)(a^2 + ab + b^2) = (a + b)(a - b\omega)(a - b\omega^2)$.
  5. $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$.

307. Show that $x^{3p} + x^{3q + 1} + x^{3r + 2}$, where $p, q, r$ are positive integers is divisible by $x^2 + x + 1$.

Solution: $x^{3p} + x^{3q + 1} + x^{3r + 2}$ will be divisible by $x^2 + x + 1$ only if all the factors of $x^2 + x + 1$ satisfy $x^{3p} + x^{3q + 1} + x^{3r + 2}$.

$x^2 + x + 1 = 0\Rightarrow x = \omega, \omega^2$. If $x = \omega$ then $x^{3p} + x^{3q + 1} + x^{3r + 2} = (\omega^3)^p + (\omega^3)^q.\omega + (\omega^3)^r.\omega^2 = 1 + \omega + \omega^2 = 0$.

If $x = \omega^2$ then $x^{3p} + x^{3q + 1} + x^{3r + 2} = (\omega^6)p + (\omega^6)^q.\omega^2 + (\omega^6)^r.\omega^4 = 1 + \omega^2 + \omega = 0$. Hence proved.

308. Show that $x^{4p} + x^{4q + 1} + x^{4r + 2} + x^{4s + 3}$, where $p, q, r, s$ are positive integers is divisible by $x^3 + x^2 + x + 1$.

Solution: Following like previous problem $x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1) = 0 \Rightarrow x = -1, \pm i$.

If $x = -1$ then $x^{4p} + x^{4q + 1} + x^{4r + 2} + x^{4s + 3} = (-1)^{4p} + (-1)^{4q + 1} + (-1)^{4r + 2} + (-1)^{4s + 3} = 1 - 1 + 1 - 1 = 0$.

If $x = i$, then $x^{4p} + x^{4q + 1} + x^{4r + 2} + x^{4s + 3} = i^{4p} + i^{4q + 1} + i^{4r + 2} + i^{4s + 3} = 1 + i - 1 - i = 0$.

If $x = -i$, then $x^{4p} + x^{4q + 1} + x^{4r + 2} + x^{4s + 3} = (-i)^{4p} + (-i)^{4q + 1} + (-i)^{4r + 2} + (-i)^{4s + 3} = 1 - i - 1 + i = 0$. Hence proved.

309. If $p = a + b + c, q = a + b\omega + c\omega^2, r = a + b\omega^2 + c\omega,$ where $\omega$ is a cube root of unity, prove that $p^3 + q^3 + r^3 - 3pqr = 27abc$.

Solution: $p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp) = (p + q + r)(p + q\omega + r\omega^2)(p + q\omega^2 + r\omega)$

$p + q + r = 3a + b(1 + \omega + \omega^2) + c(1 + \omega^2 + \omega) = 3a$. Similarly, $p + q\omega + r\omega^2 = 3c$ and $p + q\omega^2 + r\omega = 3b$. Hence, $p^3 + q^3 + r^3 - 3pqr = 27abc$, proved.

310. If $\omega$ is a cube root of unity, prove that $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3 = (2a - b - c)(2b - a - c)(2c - a - b)$.

Solution: Let $p = (a + b\omega + c\omega^2)$ and $q = (a + b\omega^2 + c\omega)$ then we know that $p^3 + q^3 = (p + q)(p + q\omega)(p + \omega^2)$.

$p + q = 2a - b - c, p + q\omega = 2b - c - a, p + q\omega^2 = 2c - a - b$, and hence $(a + b\omega + c\omega^2)^3 + (a + b\omega^2 + c\omega)^3 = (2a - b - c)(2b - a - c)(2c - a - b)$.