For the complex number $\displaystyle z = 5 \cos\left(\frac{\pi}{13}\right) + 5i \sin\left(\frac{\pi}{13}\right)$, find the complex number $\displaystyle z^{4}$ and $\displaystyle z^{-2}$.
This complex number is basically given in polar form. What does this look like in exponential form?
The number we have here is $z = 5e^{i\frac{\pi}{13}}$.
This means that the powers can be computed by $z^n = 5^n e^{i\frac{n\pi}{13}}$.
$\displaystyle z^4 = 625\cos\left(\frac{4\pi}{13}\right) + 625i \sin\left(\frac{4\pi}{13}\right)$ and $\displaystyle z^{-2} = \frac{1}{25}\cos\left(\frac{2\pi}{13}\right) - \frac{1}{25}i \sin\left(\frac{2\pi}{13}\right)$
Find the three complex solutions to the equation $\displaystyle z^3 = 1-i$.
The first step in this process is converting $1-i$ into exponential form.
For this, we get that $\displaystyle 1-i = \sqrt{2}e^{i\frac{7\pi}{4}}$.
Therefore, the three solutions here are $\displaystyle z = 2^{1/6}e^{i\frac{7\pi}{12}}$ and then two more solutions where the angle is shifted by $2\pi$ before dividing by 3.
The three solutions are $\displaystyle 2^{1/6}e^{i\frac{7\pi}{12}}$, $\displaystyle 2^{1/6}e^{i\frac{15\pi}{12}}$ and $\displaystyle 2^{1/6}e^{i\frac{23\pi}{12}}$.
Use De Moivres Formulas to find an expression for $\cos(3\theta)$ in terms of $\cos(\theta)$ and $\sin(\theta)$
De Moivres Formulas tell us that $e^{in\theta} = (\cos(\theta) + i\sin(\theta))^n$. What does this mean for $n=3$?
For $n=3$, this says that $\displaystyle \cos(3\theta) + i\sin(3\theta) = (\cos(\theta) + i\sin(3\theta))^3$.
Expand out the right-hand side, and then look at the real part of the equation. What does this tell you?
The right-hand side expands to $\displaystyle \cos^3{\theta} + 3i\cos^2{\theta} \sin{\theta} + 3i^2 \cos(\theta) \sin^2(\theta) + i^3 \sin^3(\theta)$. Using the fact that $i^2 = -1$, this can be simplified to $\displaystyle \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta) + i(3\cos^2(\theta) \sin(\theta) - \sin^3(\theta))$.
$\displaystyle \cos(3\theta) = \cos^3(\theta) - 3\cos(\theta)\sin^2(\theta)$
Find the exponential forms of the complex numbers $z_1 z_2$ and $\displaystyle \frac{z_1}{z_2}$ for the numbers $\displaystyle z_1 = \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)$ and $\displaystyle z_2 = \sqrt{5} \left(\cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right)\right)$
To find products and quotients, it is easier to convert these numbers into exponential form. What do we get for these two numbers?
We get that $\displaystyle z_1 = e^{-i\frac{\pi}{4}}$ and $z_2 = \sqrt{5}e^{\frac{\pi}{2}}$.
$\displaystyle z_1 z_2 = \sqrt{5}e^{i\frac{\pi}{4}}$ and $\displaystyle \frac{z_1}{z_2} = \frac{1}{\sqrt{5}} e^{-i\frac{3\pi}{4}}$
Use the fact that the polynomial $x^3 - 4x^2 + x + 26$ has a root at $x=-2$ to compute all three complex roots of this polynomial.
Since we know this has a root at $x=-2$, we know that $x+2$ is a factor of this polynomial. So we can use long division to get the other factor.
We see that $x^3 - 4x^4 + x + 26 = (x+2)(x^2 - 6x + 13)$.
We can find the roots of that quadratic polynomial using the quadratic formula.
The three zeros are $z=-2$, $z = 3+2i$ and $z = 3-2i$.
Find all of the 5th roots of the complex number $-4 + 4i$.
Find all roots of the equation $z^4 - 2z^3 - 2z^2 + 8 = 0$, knowing the fact that $-1+i$ is a root of this polynomial.