$(\sqrt{3} + i)^{50}$ in exponential and cartesian form

Question

Having trouble understanding the solution to the following question:

Put $(\sqrt{3} + i)^{50}$ in exponential and cartesian form.

I know the answer cartesian form is:

$$\frac{2^{50}}{2} + \frac{2^{50}\sqrt{3}}{2}i$$

And the exponential form is:

$$2^{50}e^{\frac{50\pi}{6}i}$$

But I don't know how to get there.

I know that $i^{50} = i^{4(12)+2} = i^2 = -1$. Is that somehow used in the process? What form is it given in?

Answer 1

Use the fact that$$\sqrt3+i=2\left(\frac{\sqrt3}2+\frac i2\right)=2\left(\cos\left(\frac\pi6\right)+i\sin\left(\frac\pi6\right)\right),$$together with de Moivre's formula.

Answer 2

$$e^{50i\pi/6}=e^{25i\pi /3}=e^{8i\pi+i\pi3}=(e^{2i\pi })^4e^{i\pi/3}=?$$

Answer 3

We have that

$$\sqrt 3+i=2e^{\frac{\pi}6i}\implies (\sqrt 3+i)^n=(2e^{\frac{\pi}6i})^n=2^ne^{\frac{n\pi}6i}$$

Answer 4

Exponential form is the easiest.

$\sqrt{3} + i = r*e^{i\theta}(\cos \theta + i \sin \theta)$ where $r = \sqrt{\sqrt{3}^2 + 1^2} = 2$ and $\cos \theta = \frac {\sqrt{3}}2$ and $\sin \theta = \frac 12$ So $\theta = \frac {\pi}2$.

So $(\sqrt{3}+i)^{50} = 2^{50}e^{i\frac {50\pi}6}=2^{50}e^{i(8\pi + \frac 13\pi)}=2^{50}e^{i\frac {\pi}3}$

$= 2^{50}(\cos \frac {\pi}3 + i\sin \frac {\pi}3) = 2^{49} + i2^{49}\sqrt{3}$.

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Alternatively you can try a few powers and see if patterns emerge.

$(\sqrt 3 + i)^2 = (3 -1) + 2(\sqrt 3)i = 2(1+\sqrt 3 i)$.

$(\sqrt 3 + i)^3 = 2(1 + \sqrt 3i)(\sqrt 3 +i) =2[(\sqrt 3-\sqrt 3)+ (3 + 1)i]= 8i$. (!!! single term !!!)$

$(\sqrt 3+ i)^{48}=((\sqrt 3+ i)^3)^{16}=(8i)^{16} = 8^{16}*i^{16} = (2^3)^{16}*1 = 2^{48}$

So $(\sqrt 3+i)^{50} = 2^{48}(\sqrt 3 + i)^2 = 2^{48}*2(1+\sqrt 3 i)= 2^{49} + 2^{49}\sqrt 3i$.

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It's a nice thing to learn that $[\frac 12(\sqrt {3} + i)]^3 = i$ and $[\frac 12(\sqrt {3} + i)]^{12} = 1$. So $\frac 12(\sqrt {3} + i)$ is one of the twelve twelfth roots of $1$.