Discussion questions · Section 1 of 2

A graphic calculator is not to be used in answering this question.

Solve the equation $z^2=3-4i$, giving your answers in the form $a+ib$. Hence, by expressing $\frac13i-\frac14$ in the form $k(3-4i)$, where $k$ is a constant, find the roots of the equation

$w^2=\frac13i-\frac14$.

The complex number $w$ is such that $kw^2+kww^*+iw-iw^*-1=0$, where $w^*$ is the complex conjugate of $w$ and $k$ is a real and non-zero constant.

(i) For $w=a+bi$ where $a$ and $b$ are real numbers, obtain an expression for $b$ in terms of $a$ and $k$. Explain why $w$ is either purely real or purely imaginary.

(ii) Using your result in part (i), or otherwise, find the real roots of the equation $2w^2+2ww^*+iw-iw^*-1=0$.

Solve the equation $zz^*-2z+2z^*=5-4i$.

Solve the following simultaneous equations:

(a) $4iz-3w=1+5i$ and $2z+(1+i)w=2+6i$,

(b) $iz+2w=1$ and $4z+(3-i)w^*=-6$,

giving $z$ and $w$ in the form $a+bi$ where $a$ and $b$ are real.

Consider the equation $2z^3+pz^2+qz-65=0$.

(i) If $p$ and $q$ are real numbers, what are the possibilities for real and complex roots for this equation?

It is given that $z=3+2i$ is a root of the equation $2z^3+pz^2+qz-65=0$, where $p$ and $q$ are real numbers.

(ii) Find the values of $p$ and $q$.

Do not use a calculator in answering this question.

Find the roots of the equation $w^2(1-i)+4w+(10+10i)=0$, giving your answers in cartesian form $a+ib$.