- #1
abruski
- 9
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Homework Statement
[tex]\left( \frac{2 \sqrt{3}+2i}{1-i} \right)^{50}[/tex]
Homework Equations
[tex]z = a+ib[/tex]
[tex]z = r(cos \phi + isin \phi)[/tex]
[tex]r = \sqrt{a^2+b^2}[/tex]
[tex]cos \phi = \frac{a}{r}[/tex]
[tex]sin \phi = \frac{b}{r}[/tex]
[tex]\frac{z_1}{z_2} = \frac{r_1}{r_2}\left( cos(\phi_1-\phi_2) + isin(\phi_1 - \phi_2)\right)[/tex]
[tex]z^{n} = r^{n}(cos n \phi + isin n \phi)[/tex]
The Attempt at a Solution
First I assign the numerator to [tex]z_1 = 2 \sqrt{3} + 2i[/tex] and convert it to trigonometric form:
[tex]a=2\sqrt{3} , b=2[/tex]
[tex]r = \sqrt{(2\sqrt{3})^2+2^2} = \sqrt{12+4} = 4[/tex]
[tex]cos \phi_1 = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}[/tex]
[tex]sin \phi_1 = \frac{2}{4} = \frac{1}{2}[/tex]
From the [tex]sin \phi_1[/tex] and [tex]cos \phi_1[/tex] can be concluded that that [tex]\phi_1 = \frac{\pi}{6}[/tex]. Using the conversion formula we get the trigonometric form of [tex]z_1[/tex]: [tex]z_1 = 4(cos \frac{\pi}{6} + isin \frac{\pi}{6})[/tex]
Second I assign the denominator to [tex]z_2 = 1-i[/tex]
[tex]a = 1, b = -1, r = \sqrt{2}[/tex]
[tex]cos \phi_2 = \frac{1}{\sqrt{2}}[/tex]
[tex]sin \phi_2 = -\frac{1}{\sqrt{2}}[/tex]
So [tex]\phi_2 = \frac{7\pi}{4}[/tex] and now we can derive the trigonometric form of [tex]z_2 = \sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})[/tex]
What I get is:[tex]\left( \frac{4(cos \frac{\pi}{6} + isin \frac{\pi}{6})}{\sqrt{2}(cos \frac{7\pi}{4} + isin \frac{7\pi}{4})} \right)^{50}[/tex]
Using the division formula I get the trigonometric version of the main equation:
[tex]\left( \frac{4}{\sqrt{2}}(cos(\frac{\pi}{6}-\frac{7\pi}{4}) + isin (\frac{\pi}{6}-\frac{7\pi}{4}) \right)^{50}[/tex]
How do I continue? I can simplify it further and I get [tex]cos (\frac{19\pi}{12})[/tex] and the same for sin but with the - (minus) sign in front.
I can apply the de Moivre's formula but I can't simplify it to the answer given below.
Can some one give me any guidance (step by step).
Maybe I have an error somewhere or I just don't know how to arrive to the right answer which is: [tex]2^{74}(-\sqrt{3}+i)[/tex]
Thank you in advance.
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