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Write the following numbers as an $(\alpha + \beta i)$ which means as an algebraic expression : $[2(\cos5 + i\sin5)]^{12}$ and also $(1+i)^8$ . So,as for the first one, I tried writing $2^{12}(\cos5 + i\sin5)^{12}$ but could not take it further. As for the second one, I took the modulus of $z$, which is $1$ but I don't know how to find $\alpha$ or $\beta$.

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What's for you "an algebraic expression"? Because both expressions you wrote are "algebraic", so what did you actually meant? Perhaps Polar something? – DonAntonio Dec 9 '12 at 14:26
No the first one is trigonometric algebric expression has the form : alfa +beta*i.. – einsteinwannabe Dec 9 '12 at 14:27
@Beyondhere: you are missing the point. Any number $\alpha+i\beta$ can be written using sines and cosines. The point of the question is to get rid of the power. – Martin Argerami Dec 9 '12 at 14:52

2 Answers 2

Using de Moivre's formula, $(\cos 5^{\circ}+i\sin 5^{\circ})^{12}=\cos 60^{\circ}+i\sin60^{\circ}=\frac12+i\frac{\sqrt3}2$

So, $$\{2(\cos 5^{\circ}+i\sin 5^{\circ})\}^{12}$$ $$=2^{12}(\cos 5^{\circ}+i\sin 5^{\circ})^{12}=4096\left(\frac12+i\frac{\sqrt3}2\right)=2048+i2048\sqrt3$$

$1+i=r(\cos \theta+i\sin\theta)$

Equating the real & the imaginary parts, $r\cos \theta=r\sin\theta=1$ where $r>0$

Squaring & adding we get, $r=\sqrt 2$

On division, $\tan \theta =1$ so $\theta=45^{\circ}$ as $\cos \theta,\sin\theta>0$

So, $1+i=\sqrt2(\cos45^{\circ}+i\sin45^{\circ})$

Hence, $(1+i)^8=(\sqrt2)^8\{\cos45^{\circ}+i\sin45^{\circ}\}^8=2^4(\cos360^{\circ}+i\sin360^{\circ})=16$

Alternatively, $(1+i)^2=1+i^2+2i=2i,(1+i)^4=(2i)^2=-4,(1+i)^8=(-4)^2=16$

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In the first one you have 2*(cos5..... but thank you anyway :) – einsteinwannabe Dec 9 '12 at 14:29
I've just shown the power part – lab bhattacharjee Dec 9 '12 at 14:30
If incompleteness is the reason behind the down-voting, I've just completed it. – lab bhattacharjee Dec 9 '12 at 16:40

Use the standard formula: $r(\cos\theta + i \sin\theta) = re^{i\theta}.$ From this, we can conclude that $$[r(\cos\theta + i \sin\theta)]^n = [re^{i\theta}]^n = r^ne^{in\theta} = r^n(\cos(n\theta)+i\sin(n\theta)).$$

In the first case, $r=2$, $\theta =5$ and $n=12$. Putting this into the line above:

$$[2(\cos5+i\sin5)]^{12} = 2^{12}(\cos60+i\sin60).$$

Assuming that you are working in degrees, $\cos60 = 1/2$ while $\sin 60 = \sqrt{3}/2.$ Thus:

$$[2(\cos5+i\sin5)]^{12} = 4096\left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 2048 + i \, 2048\sqrt{3} \, .$$

In the second case, note that $|1+i| = \sqrt{2}$ while $\arg(1+i) = 45^{\circ}.$ Thus:

$$1+i = \sqrt{2}(\cos45 + i \sin45) \, . $$

In this case, $r=\sqrt{2}$, $\theta = 45$ and $n = 8$, thus:

$$(1+i)^8 = 16(\cos360 + i \sin360) = 16 \, . $$

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