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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: beezy4eva on February 05, 2008, 08:01:05 pm

Title: comples number questions
Post by: beezy4eva on February 05, 2008, 08:01:05 pm: help with these questions would be appreciated :) please show working out for me :)

1. show that sinθ + cosθ i= cis (pi/2 -θ)

2. show that sinθ - cosθ i= cis (θ - pi/2)

3. simplify i(√3 -i)^7 giving the answer in modulus-Argument form.
Title: Re: comples number questions
Post by: dcc on February 05, 2008, 08:25:37 pm: Question 1:

$sin(x) + icos(x) = cis(\dfrac{\pi}{2} - x)$

$cis(\dfrac{\pi}{2} - x) = cos(\dfrac{\pi}{2} - x) + isin(\dfrac{\pi}{2} - x)$

Now we know that:

$cos(\dfrac{\pi}{2} - x) = sin(x)$ & $sin(\dfrac{\pi}{2} - x) = cos(x)$

so we can express the above as:

$cis(\dfrac{\pi}{2} - x) = sin(x) + icos(x)$

Comparing to original, we see that they are equal!

Question 2:

$sin(x) - icos(x) = cis(x - \dfrac{\pi}{2})$

$cis(x - \dfrac{\pi}{2}) = cos(x - \dfrac{\pi}{2}) + isin(x - \dfrac{\pi}{2})$

We also know that:

$cos(x - \dfrac{\pi}{2}) = sin(x)$ & $sin(x - \dfrac{\pi}{2}) = -cos(x)$

Substituting into the above:

$cis(x - \dfrac{\pi}{2}) =sin(x) - icos(x)$

Which is the same as above!

Question 3:

Simplify: $i(\sqrt{3}-i)^7$

Now, converting $z = \sqrt{3} - i$ into polar form, we get:

$|z| = \sqrt{3 + 1} = 2$

and (using the inverse tangent function i.e. tan-1(-1/sqrt(3)))

$Arg\mbox{ }z = \dfrac{-\pi}{6}$

so we can write z as such:

$z = 2cis(\dfrac{-\pi}{6})$

Now we wish to find:

$i(z^7)$ so firstly:

using de Moivre's theorem:

$z^7 = (2^7)cis(\dfrac{-7\pi}{6})$

$z^7 = 128cis(\dfrac{5\pi}{6})$ (we must convert the angle as the principal argument of Z should be within $(-\pi, \pi]$ )

converting this to cartesian co-ordinates, we get:

$z^7 = 128cos(\dfrac{5\pi}{6}) + 128isin(\dfrac{5\pi}{6})$

$z^7 = -64\sqrt{3} + 64i$

Now we are looking for i(z^7) so multiplying through by i:

$i(z^7) = -64\sqrt{3}i - 64$

$i(z^7) = -64 - 64\sqrt{3}i$

Let b be i(z^7)

$|b| = 128$

$Arg\mbox{ }b = \dfrac{-2\pi}{3}$

$b = 128cis(-\dfrac{2\pi}{3})$
Title: Re: comples number questions
Post by: beezy4eva on February 05, 2008, 08:59:27 pm: thanks a tonne :)