• Oct 31st 2007, 09:08 PM
softdrink_jr
i have 3 questions if anyone can help it would be greatly appreciated.

1) Have to use de Moivre's theorem to calculate
(1+√3i)3 / (1-i) 4
Expressing the answer in z=a+bi form.

I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!

2) If z= cosŲ + i sinŲ, show that
cos nŲ = ½ (zn + 1/n)
sin nŲ = ½i (zn + 1/zn)

3) Prove that sin5Ų + sinŲ = 2 sin 3Ų cosŲ

with the other two question, i pretty much don't have a clue on what to do!!!
• Oct 31st 2007, 10:41 PM
kalagota
Quote:

Originally Posted by softdrink_jr
i have 3 questions if anyone can help it would be greatly appreciated.

1) Have to use de Moivre's theorem to calculate
(1+√3i)3 / (1-i) 4
Expressing the answer in z=a+bi form.

I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!

so you have

$\displaystyle \frac{(1 + \sqrt3 i)^3}{(1 - i)^4}$

can you convert it to the form $\displaystyle z=r \, cis\theta$

i mean the numerator without the power and the denominator also without the power, then apply de moivre's theorem..
for the numerator part:
$\displaystyle \theta = \frac{\pi}{3}$ and r=2
for the denominator:
$\displaystyle \theta = \frac{7\pi}{4} \, \, and \, \, r=\sqrt 2$

then apply the theorem on each:
$\displaystyle (r cis\theta)^n = r^n cis n\theta$
also, apply the laws of addition and multiplication/division of complex numbers..

Quote:

Originally Posted by softdrink_jr
2) If z= cosŲ + i sinŲ, show that
cos nŲ = ½ (zn + 1/n)
sin nŲ = ½i (zn + 1/zn)

hmm, is that z^n

Quote:

Originally Posted by softdrink_jr
3) Prove that sin5Ų + sinŲ = 2 sin 3Ų cosŲ

with the other two question, i pretty much don't have a clue on what to do!!!

$\displaystyle \sin 5\theta + \sin \theta = \sin (4\theta + \theta) + \sin \theta$
$\displaystyle = \sin 4\theta \cos \theta + \sin \theta \cos 4\theta + \sin \theta$
$\displaystyle = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (\cos^2 2\theta - \sin^2 2\theta) + \sin \theta$
$\displaystyle = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (\cos^2 2\theta - \sin^2 2\theta + 1)$
$\displaystyle = 2\sin 2\theta \cos 2\theta \cos \theta + \sin \theta (2\cos^2 2\theta)$
$\displaystyle = 2\sin 2\theta \cos 2\theta \cos \theta + 2\sin \theta \cos^2 2\theta$
$\displaystyle = 2\cos 2\theta (\sin 2\theta \cos \theta + \sin \theta \cos 2\theta)$
$\displaystyle = 2\cos 2\theta \sin 3\theta$

aww, i got i wrong conclusion..
• Nov 1st 2007, 01:56 AM
softdrink_jr
correction
hmm, is that z^n

yeh thats what its meant to be z^n!!!
• Nov 1st 2007, 01:59 AM
softdrink_jr
Can u please show how to work it out proply using the laws???
[quote=kalagota;81295]so you have

$\displaystyle \frac{(1 + \sqrt3 i)^3}{(1 - i)^4}$

can you convert it to the form $\displaystyle z=r \, cis\theta$

i mean the numerator without the power and the denominator also without the power, then apply de moivre's theorem..
for the numerator part:
$\displaystyle \theta = \frac{\pi}{3}$ and r=2
for the denominator:
$\displaystyle \theta = \frac{7\pi}{4} \, \, and \, \, r=\sqrt 2$

then apply the theorem on each:
$\displaystyle (r cis\theta)^n = r^n cis n\theta$
also, apply the laws of addition and multiplication/division of complex numbers..
• Nov 1st 2007, 04:27 AM
kalagota
Quote:

Originally Posted by softdrink_jr
i have 3 questions if anyone can help it would be greatly appreciated.

1) Have to use de Moivre's theorem to calculate
(1+√3i)3 / (1-i) 4
Expressing the answer in z=a+bi form.

I have found both the polar forms for them which could possibly be wrong, its just that get confused when i convert it to a cartesian. I need to calrify my first answer and see how to do it correctly!

okay, so
numerator:
$\displaystyle z_1 = x + yi = 1 + \sqrt3 i$
$\displaystyle r = \sqrt{x^2 + y^2} = 2$
$\displaystyle \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} \sqrt3$
$\displaystyle \theta = \frac{\pi}{3} \,$ (why?)
therefore
$\displaystyle z_1 = 2 \, cis \frac{\pi}{3}$
so that (by De Moivre's)
$\displaystyle (z_1)^3 = 8 \, cis \pi$

denominator:
$\displaystyle z_2 = x + yi = 1 - i$
$\displaystyle r = \sqrt{x^2 + y^2} = \sqrt2$
$\displaystyle \theta = \tan^{-1} \frac{y}{x} = \tan^{-1} (-1)$
$\displaystyle \theta = \frac{-\pi}{4} \,$ (why?)
therefore
$\displaystyle z_2 = \sqrt2 \, cis {(-\pi)}$
so that (by De Moivre's)
$\displaystyle (z_2)^4 = 4 \, cis (-4\pi)$

now

$\displaystyle \frac{(z_1)^3}{(z_2)^4} = \frac{8 \, cis \pi}{4 \, cis -4\pi} = \frac{2(\cos \pi + i \sin \pi)}{\cos (-4\pi) + i\sin (-4\pi)} = \frac{2(-1 + i (0))}{1 + 0} = -2$

have you checked number 3? i had a different conclusion..