Solve $z^2 + 4 = 0$.
$z = \pm 2i$
Solve $z^2 - 4z + 13 = 0$. $z = 2 \pm 3i$
Given that $z = 1 + 2i$ is a root of $z^2 + bz + c = 0$ where $b, c \in \mathbb{R}$, find $b$ and $c$. $b = -2$, $c = 5$
Given that $z = 3 - i$ is a root of $z^2 + pz + q = 0$ where $p, q \in \mathbb{R}$, find $p$ and $q$. $p = -6$, $q = 10$
Show that $z = 2 + i$ is a root of $z^3 - 6z^2 + 13z - 10 = 0$, and hence find all three roots. $z = 2 + i$, $z = 2 - i$, $z = 2$
The equation $z^4 - 2z^3 + 6z^2 - 2z + 5 = 0$ has a root $z = i$. Find all four roots. $z = \pm i$, $z = 1 \pm 2i$
Find all the roots of $z^3 - 1 = 0$, expressing non-real roots in the form $re^{i\theta}$. $z = 1, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}}$
Find all six roots of $z^6 = 64$, expressing each in the form $re^{i\theta}$. $z = 2e^{i\frac{k\pi}{3}}$ for $k = -2, -1, 0, 1, 2, 3$
Find all four roots of $z^4 = -16$, expressing each in the form $a + bi$. $z = \sqrt{2}(1+i)$, $\sqrt{2}(-1+i)$, $\sqrt{2}(-1-i)$, $\sqrt{2}(1-i)$
Find the three cube roots of $8i$, giving your answers in the form $re^{i\theta}$ where $-\pi < \theta \leq \pi$. $2e^{i\frac{\pi}{6}}$, $2e^{-i\frac{\pi}{2}}$, $2e^{i\frac{5\pi}{6}}$
The equation $2z^3 - 5z^2 + 8z - 5 = 0$ has a root $z = 1$. Find the other two roots. $z = \dfrac{3}{4} \pm \dfrac{\sqrt{31}}{4}i$
Given $\omega = e^{i\frac{2\pi}{3}}$. - Show that $\omega^3 = 1$.
- Show that $1 + \omega + \omega^2 = 0$.
- Hence evaluate $(1 + \omega)(1 + \omega^2)$.
- $e^{i2\pi} = \cos 2\pi + i\sin 2\pi = 1$
- The roots of $z^3 - 1 = 0$ satisfy $z^2 + z + 1 = 0$, so $1 + \omega + \omega^2 = 0$
- $(1+\omega)(1+\omega^2) = 1 + \omega + \omega^2 + \omega^3 = 0 + 1 = 1$
- Write $2 + 2i$ in modulus-argument form.
- Find all three roots of $z^3 = 2 + 2i$, giving your answers in exact modulus-argument form.
- Show the three roots on an Argand diagram.
- $2 + 2i = 2\sqrt{2}\, e^{i\frac{\pi}{4}}$
- $z = 2^{\frac{1}{2}}e^{i\frac{\pi}{12}}$, $\;z = 2^{\frac{1}{2}}e^{i\frac{3\pi}{4}}$, $\;z = 2^{\frac{1}{2}}e^{-i\frac{7\pi}{12}}$
- Three points equally spaced at $\dfrac{2\pi}{3}$ on a circle of radius $\sqrt{2}$, starting at argument $\dfrac{\pi}{12}$
$f(z) = z^4 - 4z^3 + 14z^2 - 4z + 13$. - Given that $z = 2 + 3i$ is a root of $f(z) = 0$, write down another root.
- Find the quadratic factor of $f(z)$ corresponding to these two roots.
- Hence fully factorise $f(z)$ and find all four roots.
- $z = 2 - 3i$
- $(z - (2+3i))(z-(2-3i)) = z^2 - 4z + 13$
- $f(z) = (z^2 - 4z + 13)(z^2 + 1)$; roots: $z = 2 \pm 3i$ and $z = \pm i$