Complex numbers

Question code: 13A41

Question 1306

The polynomial

{P(z)}

has real coefficients. The equation

{P(z)=0}

has a root

{r \mathrm{e}^{\mathrm{i} \theta},}

where

{r > 0}

and

{-\pi < \theta < \pi,}

{\theta \neq 0.}

Write down a second root in terms of

{r}

and

{\theta,}

and hence show that a quadratic factor of

{P(z)}

{z^2 - 2 r z \cos \theta + r^2.}

Let

{w = 3 \mathrm{e}^{- \frac{1}{4} \pi \mathrm{i}}.}

Show that

{w^4 = -k,}

where

{k}

is a positive real number to be determined.

It is given that

{3 \mathrm{e}^{\frac{3}{4} \pi \mathrm{i}}}

is a root of

{z^4 = -81.}

Use this information and parts (i) and (ii) to write

{z^4 + 81}

as a product of two quadratic factors with real coefficients, giving each factor in non-trigonometric form.

Answer

{k = 81.}

{z^4 + 81} = {(z^2 - 3 \sqrt{2} z + 9)}\allowbreak{(z^2 + 3 \sqrt{2} z + 9)}{}.

Question code: 13A41