Complex Number Help (1 Viewer)

Kingom · Nov 29, 2016

please help. thanks

pikachu975 · Nov 29, 2016

i)

w^3 = 1
w^3 - 1 = 0
(w-1)(w^2 + w + 1) = 0
And we know that w can't equal 1, so w-1 can't equal 0. This means that w^2 + w + 1 must equal 0.
Therefore w^2 + w + 1 = 0 QED

ii)

not sure

iii)

Sub in z into the equation then realise the denominator.

fluffchuck · Nov 30, 2016

ii) Common denominate everything and expand. Make sure you use the conversion 1+w+w^2=0 and w^3=1

Mahan1 · Dec 3, 2016

ii) First method, start from LHS:
$\frac{1}{z-1} + \frac{1}{\omega z -1} + \frac{1}{\omega^{2} z-1}$

$\frac{1}{z-1} + \frac{1}{\omega (z -\omega^{2})} + \frac{1}{\omega^{2} ( z-\omega)}$

$\frac{1}{z-1} + \frac{\omega^{2}}{ (z -\omega^{2})} + \frac{\omega}{ ( z-\omega)} \ \ \ \textrm{using the fact} \ \ \omega = \frac{1}{\omega^{2}}, \omega^{2} = \frac{1}{\omega}$

$\frac{(z- \omega^{2}) (z - \omega)}{z^{3}-1} + \frac{\omega^{2}(z-1)(z- \omega)}{z^{3}-1} + \frac{\omega (z-1)(z- \omega^{2})}{ z^{3} -1}$

$\textrm{since} \ (z-1)(z- \omega)(z - \omega^{2}) = z^{3} -1$

$= \frac{3}{z^{3}-1}$

Second method, starting from RHS:

$\frac{3}{z^{3} -1}$

$\frac{3}{z^{3} -\omega^{3}}$

$\frac{3}{(z-\omega)(z^{2} + \omega z + \omega^{2})}$

$\frac{3}{(z-\omega)(z^{2} + \omega z -1 - \omega)}$

$\frac{3}{(z-\omega)(z-1)(z+1+ \omega)}$

$\frac{3}{(z-\omega)(z-1)(z - \omega^{2})}$

$\textrm{using partial fraction} \ \ \ \ \frac{1}{z-1} + \frac{\omega}{z - \omega} + \frac{\omega^{2}}{z - \omega^{2}} \ \ \ \textrm{this is equivalent to the question}$

iv) Use the hint, part iii) and equate the imaginary parts of the equation in part ii)

Complex Number Help (1 Viewer)

Kingom

Member

Attachments

pikachu975

Premium Member

fluffchuck

Active Member

Mahan1

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)