25. Complex Numbers Problems Part 5#
Prove that
\[\left(\frac{1 + sin\phi + icos\phi}{1 + sin\phi - icos\phi}\right)^n = cos\left(\frac{n\pi}{2} + n\phi\right) + isin\left(\frac{n\pi}{2} - n\phi\right) \]If \(sin\alpha + sin\beta + sin\gamma = cos\alpha + cos\beta + cos\gamma = 0,\) show that \(cos3\alpha + cos3\beta + cos3\gamma = 3cos(\alpha + \beta + \gamma)\) and \(sin3\alpha + sin3\beta + sin3\gamma = 3sin(\alpha + \beta + \gamma)\)
If \(sin\alpha + sin\beta + sin\gamma = cos\alpha + cos\beta + cos\gamma = 0,\) show that \(cos2\alpha + cos2\beta + cos2\gamma = sin2\alpha + sin2\beta + sin2\gamma = 0\)
If \(\alpha, \beta\) are the roots of the equation \(t^2 - 2t + 2 = 0,\) show that a value of \(x,\) satisfying
\[\frac{(x + \alpha)^n - (x + \beta)^b}{(\alpha - \beta)} = \frac{sin\theta}{sin^n\theta} \text{ is } x = cot\theta - 1 \]If \((1 + x)^n = p_0 + p_1x + p_2x^2 + ... + p_nx^n,\) show that
\[p_0 - p_2 + p_4 ... = 2^{\frac{n}{2}}cos\frac{n\pi}{4} \text{ and } p_1 - p_3 + p_5 + ... = 2^{\frac{n}{2}}sin\frac{n\pi}{4}\]If \((1 - x + x^2)^n = a_0 + a_1 + a_2x^2 + ... a_{2n}x^{2n}\) show that
\[a_0 + a_3 + a_6 + ... = \frac{1}{3}\left(1 + 2^{n + 1}cos\frac{n\pi}{3}\right) \]If \(n\) is a positive integer and \((1 + x)^n = c_0 + c_1x + c_2x^2 + ... + c_nx^n,\) show that
\[c_0 + c_4 + c_8 + ... = 2^{n - 2} + 2^{\frac{n}{2} - 1}cos\frac{n\pi}{4}. \]Solve the equation \(z^8 + 1 = 0\) and deduce that
\[cos4\theta = 8\left(cos\theta - cos\frac{\pi}{8}\right)\left(cos\theta - cos\frac{3\pi}{8}\right)\left(cos\theta - cos\frac{5\pi}{8}\right)\left(cos\theta - cos\frac{7\pi}{8}\right) \]Prove that the roots of the equation \(8x^3 - 4x^2 - 4x + 1 = 0\) are \(cos\frac{\pi}{7}, cos\frac{3\pi}{7}, cos\frac{5\pi}{7}.\)
Solve the equation \(z^{10} - 1 = 0\) and deduce that
\[sin5\theta = 5sin\theta\left(1 - \frac{sin\theta}{sin^2\frac{\pi}{5}}\right)\left(1 - \frac{sin\theta}{sin^2\frac{2\pi}{5}}\right) \]Solve the equation \(x^7 + 1 = 0\) and deduce that
\[cos\frac{\pi}{7}cos\frac{3\pi}{7}cos\frac{5\pi}{7} = -\frac{1}{8} \]Form the equation whose roots are \(cot^2\frac{\pi}{2n + 1}, cot^2\frac{2\pi}{2n + 1}, ..., cot^2\frac{n\pi}{2n + 1}\) and hence find the value of \(cot^2\frac{\pi}{2n + 1} + cot^2\frac{2\pi}{2n + 1} + ... + cot^2\frac{n\pi}{2n + 1}\)
If \(\theta \ne k\pi,\) show that \(cos\theta sin\theta + cos^2\theta sin2\theta + ... +cos^n\theta sinn\theta = cot\theta(1 - cos^n\theta cosn\theta)\)
Show that \(-3 -4i = 5e^{i\left(\pi + tan^{-1}\frac{4}{3}\right)}\)
Solve the equation \(2\sqrt{2}x^4 = (\sqrt{3} - 1) + i(\sqrt{3} + 1)\)
Show that \(\left(\frac{1 + cos\phi + isin\phi}{1 + cos\phi -isin\phi}\right)^n = cosn\phi + isinn\phi\)
If \(2cos\theta = x + \frac{1}{x}\) and \(2cos\phi = y + \frac{1}{y},\) then prove that
\(\frac{x}{y} + \frac{y}{x} = 2cos(\theta - \phi)\)
\(xy + \frac{1}{xy} = 2cos(\theta + \phi)\)
\(x^my^n + \frac{1}{x^my^n} = 2cos(m\theta + n\phi)\)
\(\frac{x^m}{y^n} + \frac{y^n}{x^m} = 2cos(m\theta - n\phi)\)
If \(\alpha, \beta\) are the roots of the equation \(x^2 -2x +4 = 0,\) prove that \(\alpha^n + \beta^n = 2^{n + 1}cos\frac{n\pi}{3}\)
Find the equation whose roots are \(n\) th powers of the roots of the equation \(x^2 -2xcos\theta + 1 = 0\)
If \(\alpha, \beta\) are imaginary cube roots of 1 then show that
\[\alpha e^{\alpha x} + \beta e^{\beta x} = -e^\frac{x}{2}\left[cos\left(\frac{\sqrt{3}}{2}x\right) + \sqrt{3}\left(\frac{\sqrt{3}}{2}x\right)\right] \]Find the values of \(A\) and \(B\) where \(Ae^{2i\theta} + Be^{-2i\theta} = 5cos2\theta - 7sin2\theta\)
If \(x = cos\theta + isin\theta\) and \(\sqrt{1 - c^2} = nc - 1,\) prove that
\[(1 + c cos\theta) = \frac{c}{2n}(1 + nx)\left(1 + \frac{n}{x}\right) \]Show that the roots of equation \((1 + z)^n = (1 -z)^n\) are \(itan\frac{r\pi}{n}, r = 0, 1, 2, ..., (n - 1)\) excluding the value when \(n\) is even and \(r = \frac{n}{2}.\)
If \(x = cos\alpha + isin\alpha, y = cos\beta + isin\beta,\) show that
\[\frac{(x + y)(xy - 1)}{(x - y)(xy + 1)} = \frac{sin\alpha + sin\beta}{sin\alpha - sin\beta} \]
Since we have not covered permutations and combinations let me give the formulas:
Show that
\[^nC_0 + {^nC_3} + {^nC_6} + ... = \frac{1}{3}\left[2^n + 2cos\frac{n\pi}{3}\right] \]Show that
\[^nC_1 + {^nC_4} + {^nC_7} + ... = \frac{1}{3}\left[2^{n - 2} + 2cos\frac{(n - 2)\pi}{3}\right] \]Show that
\[^nC_2 + {^nC_5} + {^nC_8} + ... = \frac{1}{3}\left[2^{n + 2} + 2cos\frac{(n + 2)\pi}{3}\right] \]If \((1 - x + x^2)^{6n} = a_0 + a_1x + a_2x^2 + ...,\) show that
\[a_0 + a_3 + a_6 + ... = \frac{1}{3}(2^{6n + 1} + 1) \]If \((1 - x + x^2)^{n} = a_0 + a_1x + a_2x^2 + ...,\) show that
\[a_0 + a_3 + a_6 + ... = \frac{1}{3}(1 + 2^{n + 1} cos\frac{n\pi}{3}) \]Let
\[A = x + y +z, A' = x' + y' + z', AA' = x'' + y'' + z'', B = x + y\omega + z\omega^2, B' = x' + y'\omega + z'\omega^2, BB' = x'' + y''\omega + z''\omega^2, C = x + y\omega^2 + z\omega, C' = x' + y'\omega^2 + z'\omega, CC' = x'' + y''\omega^2 + z''\omega.\]then find \(x'', y''\) and \(z''\) in terms of \(x, y, z\) and \(x', y', z'.\)
Prove the equality
\[(ax - by -cz -dt)^2 + (bx + ay -dz + ct)^2 + (cx + dy + az -bt)^2 + (dx - cy + bz + at)^2 = \]\[(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + z^2 + t^2). \]Prove the following equalities:
\[\frac{cosn\theta}{cos^n\theta} = 1 - {^nC_2}tan^2\theta + {^nC_4}tan^4\theta - ... + A \text{ where } A = (-1)^\frac{n}{2}~tan^n\theta \text{ if } n \text{ is even,} A = (-1)^\frac{n - 1}{2}~{^nC_{n - 1}}tan^n\theta \text{ if } n \text{ is odd;} \frac{sinn\theta}{cos^n\theta} = {^nC_1}tan\theta - {^nC_3}tan^3\theta + {^nC_5}tan^5\theta - ... + A \text{ where } A = (-1)^\frac{n - 2}{2}~{^nC_{n - 1}}tan^{n - 1}\theta \text{ if } n \text{ is odd, } A = (-1)^\frac{n}{2}~tan^n\theta \text{ if } n \text{ is odd.}\]Prove the following equality:
\[2^{2m}cos^{2m}x = \sum_{k = 0}^{k = m - 1} 2 {2m \choose k} cos2(m - k)x + {2m \choose m} \]Prove the following equality:
\[2^{2m}sin^{2m}x = \sum_{k = 0}^{k = m - 1} (-1)^{m + k} 2 {2m \choose k} cos2(m - k)x + {2m \choose m} \]Prove the following equality:
\[2^{2m}cos^{2m + 1}x = \sum_{k = 0}^{k = m} 2 {{2m + 1} \choose k} cos(2m - 2k + 1)x \]Prove the following equality:
\[2^{2m}sin^{2m + 1}x = \sum_{k = 0}^{k = m} (-1)^{m + k} 2 {{2m + 1} \choose k} cos(2m - 2k + 1)x \]Let
\[u_n = cos\alpha + r cos(\alpha + \theta) + r^2 cos(\alpha +2\theta) + ... + r^n cos(\alpha + n\theta) v_n = sin\alpha + r sin(\alpha + \theta) + r^2 sin(\alpha +2\theta) + ... + r^n sin(\alpha + n\theta)\]then show that
\[u_n = \frac{cos\alpha - r cos(\alpha - \theta) - r^{n + 1} cos[(n + 1)\theta + \alpha] + r^{n + 2} cos(n\theta + \alpha)}{1 - 2rcos\theta + r^2} v_n = \frac{sin\alpha - r sin(\alpha - \theta) - r^{n + 1} sin[(n + 1)\theta + \alpha] + r^{n + 2} sin(n\theta + \alpha)}{1 - 2rcos\theta + r^2}\]Simplify the following sum:
\[S = 1 + n cos \theta + \frac{n(n - 1)}{1*2} cos2\theta + ... = \sum_{k = 0}^{k = n}C^n_k cosk\theta \]Simplify the following sum:
\[S = 1 + n sin \theta + \frac{n(n - 1)}{1*2} sin2\theta + ... = \sum_{k = 0}^{k = n}C^n_k sink\theta \]If \(\alpha = \frac{\pi}{2n}\) and \(o < 2n\) then prove that
\[sin^{2p} \alpha + sin^{2p} 2\alpha + ... + sin^{2p} n\alpha = \frac{1}{2} + n\frac{1 *3 * 5 * ... (2p - 1)}{2 * 4 * ... 2p} \]Prove that the polynomial \(x(x^{n - 1} -na^{n - 1}) + a^n(n - 1)\) is divisible by \((x - a)^2.\)
Prove that \((x + y)^n - x^n - y^n\) is divisible by \(xy(x + y)(x^2 + xy + y^2)\) if \(n\) is an odd number and not divisible by 3.
Find out whether the polynomial \(x^{4a} + x^{4b + 1} + x^{4c + 2} + x^{4d + 3}\) is divisible by \(x^3 + x^2 + x + 1\) where \(a, b, c, d\) are positive integers.
Prove that the polynomial \((cos\theta + x sin\theta)^n - \cos n\theta - x sin n\theta\) is divisible by \(x^2 + 1.\)
Prove that the polynomial \(x^n sin\theta - k^{n - 1}x sin n\theta + k^n sin(n - 1)\theta\) is divisible by \(x^2 - 2kx cos\theta + k^2.\)
Find the sum of the \(p\) the powers of the roots of the equation \(x^n - 1 = 0\) where \(p\) is a positive integer.
Let \(\alpha = cos\frac{2\pi}{n} + isin\frac{2\pi}{n}\) where \(n\) is a positive integer and let
\[A_k = x + y\alpha^k + z\alpha^{2k} + ... + w\alpha^{(n - 1)k} \text{ where, } k = 0, 1, 2, 3 ..., n - 1\]where, \(x, y, z, ..., u, w\) and \(n\) are arbitrary complex numbers.
Prove that
\[\sum_{k = 0}^{k = n - 1}|A_k|^2 = n\{|x|^2 + |y|^2 + ... + |w|^2\} \]
Prove the following identities:
- \[x^{2n} - 1= (x^2 - 1)\sum_{k = 1}^{k = n - 1}\left(x^2 - 2xcos\frac{k\pi}{n} + 1\right) \]
- \[x^{2n + 1} - 1 = (x - 1)\sum_{k = 1}^{k = n}\left(x^2 - 2xcos\frac{2k\pi}{2n + 1} + 1\right) \]
- \[x^{2n + 1} - 1= (x + 1)\sum_{k = 1}^{k = n}\left(x^2 + 2xcos\frac{2k\pi}{2n + 1} + 1\right) \]