Given complex numbers $z_1 = 3 - 4i$, and $z_2 = 2+i$, compute $2z_1 - z_2$, $z_1\cdot z_2$ and $z_2 - z_1$
For addition and subtraction, we need to add and subtract the real and imaginary parts separately.
When multiplying by a real number, we can just distribute that number to each of the real and imaginary parts.
For multiplying complex numbers, we distribute everything out, treating $i$ like a variable, and then using the fact that $i^2 = -1$ to simplify the expression.
$2z_1 - z_2 = 4 - 9 i$, $z_1\cdot z_2 = 10 - 5i$ and $z_2 - z_1 = -1 + 5i$.
For $z_1 = 3 - i$ and $z_2 = 2 + 3i$, calculate $\bar{z_1}$, $\bar{z_2}$, $\displaystyle \frac{1}{z_1}$ and $\displaystyle \frac{z_2}{z_1}$
To find the complex conjugate, we need to switch the sign of the imaginary part.
For any complex number $\displaystyle \frac{1}{z} = \frac{\bar{z}}{|z|^2}$.
Once you get $\frac{1}{z}$, the quotient of two complex numbers can be found by multiplying them.
For $z_1$, we get that $\displaystyle \frac{1}{z_1} = \frac{3+i}{10} = \frac{3}{10} + \frac{1}{10}i$
$\bar{z_1} = 3+i$, $\bar{z_2} = 2-3i$, $\displaystyle \frac{1}{z_1}= \frac{3}{10} + \frac{1}{10}i$ and $\displaystyle \frac{z_2}{z_1}= \frac{3}{10} + \frac{11}{10}i$
Convert the complex number $z = -2\sqrt{3} + 2i$ into exponential form, and find the exponential form of $\bar{z}$.
We need to find $r$ and $\theta$ for the point $(-2\sqrt{3}, 2)$ in Cartesian coordinates.
We get that $r^2 = (-2\sqrt{3})^2 + 2^2 = 16$ and $\displaystyle \theta = \tan^{-1}\left(\frac{2}{-2\sqrt{3}}\right) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
So $r = 4$ and $\displaystyle \theta = -\frac{\pi}{6}$, but this $\theta$ is in the wrong quadrant. So we actually want $\displaystyle \frac{5\pi}{6}$ for $\theta$.
To get $\bar{z}$, we need to reflect over the $x$ axis. This results in taking the negative of the $\theta$ angle. We can also just start with $\bar{z} = -2\sqrt{3} - 2i $ and go from there.
$\displaystyle z = 4e^{i\frac{5\pi}{6}}$, $\displaystyle \bar{z} = 4e^{-i\frac{5\pi}{6}} = 4e^{i\frac{7\pi}{6}}$
Find the partial fraction decomposition of $\displaystyle \frac{6x^4 - 13x^3 + 37 x^2 - 37 x - 1}{(x - 2) (x^2 + 1) (x^2 + 9)}$.
The decomposition here looks like $\frac{A}{x-2} + \frac{Bx + C}{x^2 + 1} + \frac{Dx + E}{x^2 + 9}$
To set up value substitution, we end up with $\displaystyle 6x^4 - 13x^3 + 37 x^2 - 37 x - 1 = A(x^2 + 1)(x^2 + 9) + (Bx + C)(x-2)(x^2 + 9) + (Dx + E)(x-2)(x^2 + 1)$.
To set this up, we need to plug in $x=2$, $x=i$, and $x=3i$.
Each of the complex numbers gives rise to a system of two equations to solve for two coefficients, with these equations coming from the real and imaginary parts of the complex numbers.
$\displaystyle \frac{1}{(x-2)} + \frac{2x+1}{x^2 +1} + \frac{3x-4}{x^2 + 9}$
Find the four complex numbers where the function $\displaystyle f(x) = \frac{1}{(x^2 + 1)(x^2 + 2x + 5)}$ does not exist, and use this to determine an upper bound on the radius of convergence of the power series expansion of $f(x)$ centered at $x=-3$.
The denominator of this function is $(x^2 + 1)(x^2 + 2x + 5)$, so we can find where this part is zero.
The roots we get here are $z_1 = i$, $z_2 = -i$, and $z_3 = -1 + 2i$, $z_4 = -1 - 2i$.
To find the radius of convergence, or an upper bound on it, we need to figure out how far each of these numbers are from $-3$, because the radius can not extend past any of these points.
The distances are $\sqrt{10}$, $\sqrt{10}$, $\sqrt{8}$ and $\sqrt{8}$.
$f(x)$ does not exist at $z_1 = i$, $z_2 = -i$, and $z_3 = -1 + 2i$, $z_4 = -1 - 2i$. An upper bound on the radius of convergence is $\sqrt{8}$.
Use complex numbers to help compute $\displaystyle \int \frac{x^2 - 5x + 9}{(x+1)(x^2 + 4)}\ dx$
Use complex numbers to find an upper bound on the radius of convergence of the power series expansion of $\frac{\sin{x}}{x^2 + 9}$ centered at $x=1$. Do not compute the power series expansion.