Mathematics Extension 2

Solve the quadratic equation $z^2-3z+4=0$, where $z$ is a complex number. Give your answers in Cartesian form.

Let $\alpha$ be a real number and suppose that $z$ is a complex number such that

You may assume that

for all positive integer $n$. Let

Hence or otherwise, find all solutions of

Let $\alpha$ be a real number and suppose that $z$ is a complex number such that

You may assume that

for all positive integer $n$. Let

Find the ninth roots of unity.

The complex number $2+i$ is a zero of $P(z)=z^4-3z^3+cz^2+dz-30$, where $c,d\in\mathbb R$. Explain why $2-i$ is also a zero of $P(z)$.

Find the cube roots of $2-2i$. Give your answer in exponential form.

By considering the roots of $z^3=4\sqrt{2}(1+i)$, show that

Use De Moivre’s Theorem to solve the equation

One of the roots of $iz^2+3z+3-11i=0$ is $3+2i$. What is the other root?

A. $-2i$
B. $-3+i$
C. $3-2i$
D. $-5-3i$

$z^2+pz+q=0$, where $p,q$ are real, has one solution $z=3-2i$. The values of $p$ and $q$ are:

A. $p=6$, $q=13$

B. $p=-6$, $q=13$

C. $p=-6$, $q=-13$

D. $p=-6$, $q=-13$

Hence solve the equation $z^2-3iz-(4+6i)=0$.

Hence, find the exact value of $(-\sqrt3+i)^{10}$, giving your answer in the form $x+iy$.

Using De Moivre’s Theorem, find the 3 cube roots of $-4\sqrt2+4i\sqrt2$. Leave your answers in modulus/argument form.

Show that if $z$ is a solution of $z^5+1=0$ and $z\ne-1$, then $u=z+\frac1z$ is a solution of $u^2-u-1=0$.

Consider the equation $z^5+1=0$, where $z$ is a complex number. Solve the equation $z^5+1=0$ by finding the fifth roots of $-1$.

Hence, find the exact value of $\cos\frac{3\pi}{5}$.

Simplify fully
$\frac1{1+\omega}+\frac1{1+\omega^2},$
given that $\omega$ is a non-real cube root of unity.

Find all the complex numbers $z_1,z_2,z_3$ that satisfy the following three conditions simultaneously:
$|z_1|=|z_2|=|z_3|,$
$z_1+z_2+z_3=1,$
$z_1z_2z_3=1.$

Find the roots of $w^5=1$. Express your answers in the form $\cos\theta+i\sin\theta$, where $-\pi<\theta\le\pi$.

Using parts (i) and (ii), find the roots of
$\left(\frac{2+z}{2-z}\right)^5=1.$

Which polynomial could have $2+i$ as a zero, given that $k$ is real?

A. $x^3-4x^2+kx$
B. $x^3-4x^2+kx+5$
C. $x^3-5x^2+kx$
D. $x^3-5x^2+kx+5$

Hence, or otherwise, solve $z^2+2z+1+i=0$, giving your solutions in the form $a+ib$ where $a$ and $b$ are real numbers.

By using part (i), or otherwise, show that
$\operatorname{Re}\left(e^{i\pi/10}\right)=\sqrt{\frac{5+\sqrt5}{8}}.$

Given that $z=3+i$ is a root of $z^2+pz+q=0$, where $p$ and $q$ are real, what are the values of $p$ and $q$?

A. $p=-6$, $q=\sqrt{10}$
B. $$, C. , D. ,

Solve $z^2+3z+(3-i)=0$, giving your answer(s) in the form $a+bi$, where $a$ and $b$ are real.