Ejercicios resueltos de potencias y raíces

Primero convertir a forma polar: $2 + 2i = 2\sqrt{2}e^{i\pi/4}$

$(2 + 2i)^4 = (2\sqrt{2})^4 e^{i4\pi/4} = 64e^{i\pi} = 64(-1) = -64$

$1 - i = \sqrt{2}e^{-i\pi/4}$

$(1 - i)^8 = (\sqrt{2})^8 e^{-i8\pi/4} = 16e^{-i2\pi} = 16e^{i0} = 16$

$-1 + \sqrt{3}i = 2e^{i2\pi/3}$

$(-1 + \sqrt{3}i)^6 = 2^6 e^{i12\pi/3} = 64e^{i4\pi} = 64e^{i0} = 64$

$\frac{1 + i}{\sqrt{2}} = \frac{\sqrt{2}e^{i\pi/4}}{\sqrt{2}} = e^{i\pi/4}$

$\left(\frac{1 + i}{\sqrt{2}}\right)^{12} = (e^{i\pi/4})^{12} = e^{i3\pi} = e^{i\pi} = -1$

$8 = 8e^{i0}$

$z_k = 8^{1/3}e^{i(0 + 2\pi k)/3} = 2e^{i2\pi k/3}$ para $k = 0, 1, 2$

$z_0 = 2e^{i0} = 2$

$z_1 = 2e^{i2\pi/3} = 2\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -1 + \sqrt{3}i$

$z_2 = 2e^{i4\pi/3} = 2\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -1 - \sqrt{3}i$

$-16 = 16e^{i\pi}$

$z_k = 16^{1/4}e^{i(\pi + 2\pi k)/4} = 2e^{i\pi(2k+1)/4}$ para $k = 0, 1, 2, 3$

$z_0 = 2e^{i\pi/4} = \sqrt{2} + \sqrt{2}i$

$z_1 = 2e^{i3\pi/4} = -\sqrt{2} + \sqrt{2}i$

$z_2 = 2e^{i5\pi/4} = -\sqrt{2} - \sqrt{2}i$

$z_3 = 2e^{i7\pi/4} = \sqrt{2} - \sqrt{2}i$

$64i = 64e^{i\pi/2}$

$z_k = 64^{1/6}e^{i(\pi/2 + 2\pi k)/6} = 2e^{i\pi(4k+1)/12}$ para $k = 0, 1, 2, 3, 4, 5$

$z_0 = 2e^{i\pi/12}$, $z_1 = 2e^{i5\pi/12}$, $z_2 = 2e^{i3\pi/4}$

$z_3 = 2e^{i13\pi/12}$, $z_4 = 2e^{i17\pi/12}$, $z_5 = 2e^{i7\pi/4}$

$z = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2}$

$z_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{i2\pi/3}$

$z_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i = e^{i4\pi/3}$

$z_k = (32)^{1/5}e^{i(\pi/3 + 2\pi k)/5} = 2e^{i\pi(6k+1)/15}$ para $k = 0, 1, 2, 3, 4$

$z_0 = 2e^{i\pi/15}$, $z_1 = 2e^{i7\pi/15}$, $z_2 = 2e^{i13\pi/15}$

$z_3 = 2e^{i19\pi/15}$, $z_4 = 2e^{i25\pi/15} = 2e^{i5\pi/3}$

$1 + i\sqrt{3} = 2e^{i\pi/3}$

$(1 + i\sqrt{3})^{10} = 2^{10}e^{i10\pi/3} = 1024e^{i(2\pi + 4\pi/3)} = 1024e^{i4\pi/3} = 1024\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -512 - 512\sqrt{3}i$

$\frac{1 - i}{1 + i} = \frac{(1 - i)(1 - i)}{(1 + i)(1 - i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{-2i}{2} = -i = e^{-i\pi/2}$

$\left(\frac{1 - i}{1 + i}\right)^{20} = (e^{-i\pi/2})^{20} = e^{-i10\pi} = e^{i0} = 1$

Como $w = e^{i2\pi/7}$, tenemos $w^7 = e^{i2\pi} = 1$, por tanto $w^7 - 1 = 0$

$(w - 1)(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) = 0$

Como $w \neq 1$, debe cumplirse: $w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 = 0$

Por tanto: $w + w^2 + w^3 + w^4 + w^5 + w^6 = -1$

Sea $S = 1 + z + z^2 + \cdots + z^{n-1}$

$zS = z + z^2 + z^3 + \cdots + z^n$

$S - zS = 1 - z^n$

$S(1 - z) = 1 - z^n$

Si $z \neq 1$: $S = \frac{1 - z^n}{1 - z} = \frac{z^n - 1}{z - 1}$ ✓

Las raíces n-ésimas de la unidad son $\omega_k = e^{i2\pi k/n}$ para $k = 0, 1, \ldots, n-1$

$\sum_{k=0}^{n-1} \omega_k = \sum_{k=0}^{n-1} e^{i2\pi k/n}$

Usando la fórmula con $z = e^{i2\pi/n}$:

$\sum_{k=0}^{n-1} z^k = \frac{z^n - 1}{z - 1} = \frac{e^{i2\pi} - 1}{e^{i2\pi/n} - 1} = \frac{1 - 1}{e^{i2\pi/n} - 1} = 0$ ✓

$\sum_{k=0}^{7} e^{ik\pi/4} = \sum_{k=0}^{7} (e^{i\pi/4})^k$

Con $z = e^{i\pi/4}$: $z^8 = e^{i2\pi} = 1$

$\sum_{k=0}^{7} z^k = \frac{z^8 - 1}{z - 1} = \frac{1 - 1}{e^{i\pi/4} - 1} = 0$

Las raíces cúbicas de la unidad son: $1$, $e^{i2\pi/3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$, $e^{i4\pi/3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$

Suma: $1 + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 1 - 1 + 0i = 0$ ✓