- $i^{4n+2} = -1$
- $i^{4n+3} = -i$
-Divide by 4 and take remainder.
+For $i^n$, divide $n$ by 4 and let remainder $= r$. Then $i^n = i^r$.
### Multiplying complex expressions
- $z \overline{z} = |z|^2$
- $z + \overline{z} = 2 \operatorname{Re}(z)$
-
### Modulus
Distance from origin.
$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$
-## Polar form
+*Must include $\pm$ in solutions*
+
+## Solving complex polynomials
+
+#### Dividing complex polynomials
+
+Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that:
+
+$$P(z) = D(z)Q(z) + R(z)$$
+
+#### Remainder theorem
-General form:
-$z=r \operatorname{cis} \theta$
-$= r(\operatorname{cos}\theta+i \operatorname{sin}\theta)$
+Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$
-$z=a+bi$
-$z=r\operatorname{cis}\theta$
+## Conjugate root theorem
+Let $P(z)$ be a polynomial with real coefficients. If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, the the conjugate $a-bi$ is also a solution.
-- $z=a+bi$
-- $r$ is the distance from origin, given by Pythagoras ($r=\sqrt{x^2+y^2}$)
-- $\theta$ is the argument of $z$, CCW from origin
+## Polar form
+
+$$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$
+
+- $r=|z|$, given by Pythagoras ($r=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$)
+- $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`)
+- **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$)
Note each complex number has multiple polar representations:
$z=r \operatorname{cis} \theta = r \operatorname{cis} (\theta+2 n\pi$) where $n$ is integer number of revolutions
$$(r \operatorname{cis} \theta)^{-1} = r\operatorname{cis} (- \theta)$$
+Reflection of $z$ across horizontal axis.
+
### Multiplication and division in polar form
$z_1z_2=r_1r_2\operatorname{cis}(\theta_1+\theta_2)$ (multiply moduli, add angles)
## de Moivres' Theorem
$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ where $n \in \mathbb{Z}$
+
+## Roots of complex numbers
+
+$n$th roots of $r \operatorname{cis} \theta$ are:
+$z={r^{1 \over n}} \cdot (\cos ({{\theta + 2k \pi} \over n}) + i \sin ({{\theta + 2 k \pi} \over n}))$
+
+Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$
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