How to raise a complex numbers to a power?

Question

$$((1+\sqrt{3}i)/2)^n$$

What is the cartesian form of it?

We never had complex numbers at university . So i have to teach myself. My problem is the exponent.

Answer 1

First note that $$\frac{1+\sqrt{3}i}{2}=\frac12+i\frac{\sqrt{3}}{2}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}=\exp\frac{i\pi}{3}.$$Hence $$\left(\frac{1+\sqrt{3}i}{2}\right)^n=\exp\frac{i\pi n}{3}=\cos\frac{\pi n}{3}+i\sin\frac{\pi n}{3}.$$

Answer 2

The first way is to compute successive powers of $z_0=\big(\frac{1+i\sqrt{3}}{2}\big)$.

$z_0^2=\big(\frac{1+i\sqrt{3}}{2}\big)^2=\frac{-1+i\sqrt{3}}{2}$ and $$z_0^3=\big(\frac{1+i\sqrt{3}}{2}\big)^3= \big(\frac{1+i\sqrt{3}}{2}\big)^2\times \big(\frac{1+i\sqrt{3}}{2}\big)= \big(\frac{-1+i\sqrt{3}}{2}\big)\times\big(\frac{1+i\sqrt{3}}{2}\big)=-1\Longrightarrow z_0^6=1$$ Thus:

$$z_0^{6n}=1$$ $$z_0^{6n+1}=z_0$$ $$ z_0^{6n+2}=z_0^2$$ $$z_0^{6n+3}=-1$$ $$z_0^{6n+4}=-z_0$$ $$z_0^{6n+5}=-z_0^2$$

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Another method is to write $z_0$ in polar. It is not hard to find that $z_0=\cos\big(\frac{\pi}{3}\big)+\sin\big(\frac{\pi}{3}\big)$. Now we use De Moivre Identity: $$ z_0^n=\bigg(\cos\big(\frac{\pi}{3}\big)+i\sin\big(\frac{\pi}{3}\big)\bigg)^n=\cos\big(\frac{n\pi}{3}\big)+i\sin\big(\frac{n\pi}{3}\big).$$ We compute the cosine and the sine modulo $6$ i.e: We write $n$ as $6p+r$ with $r\in\{0,1,...,5\}$

Answer 3

The simplest an easiest way would be to use polar form. Any complex number can be written in the form $r(cos\theta + \iota sin\theta)$ where $r\ge 0$. In polar form this can be written as $re^{\iota\theta}$. So raising it to any power n gives $r^ne^{n\iota\theta }$, which can be rewritten in the form $r^n(cos (n\theta) + \iota sin (n\theta))$

Answer 4

The "Cartesian form" of a complex number, a+ bi, corresponds to a point, (a, b) on a Cartesian coordinate system. It's "polar form", either $r cos(\theta)+i r sin(\theta)$ or $re^{i\theta}$, corresponds to a point $(r, \theta)$ in a polar coordinate system. To get the conversion formulas, draw a line from (a, b) to the origin. r is the length of that line, $r= \sqrt{a^2+ b^2}$ and $\theta$ is the angle that line makes with the positive x (real) axis, $theta= arctan\left(\frac{b}{a}\right)$. In this problem, $a= \frac{1}{2}$ and $b= \frac{\sqrt{3}}{2}$ so $r= \sqrt{\frac{1}{4}+ \frac{3}{4}}= 1$ and $\theta= arctan\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right)= arctan\left(\sqrt{3}\right)= \frac{\pi}{3}$

Answer 5

Ignoring the polar method, just expand using the binomial formula,

$$(1+i\sqrt3)^n=\sum_{k=0}^n\binom nki^k\sqrt3^k.$$

Note that for even $k$, the factors will be $3^{k/2}$ with alternating signs, and for odd $k$, $i\sqrt3\,3^{(k-1)/2}$, also with alternating signs.

In the end,

$$\frac{p+i\sqrt3 q}{2^n}$$ where $p,q$ are some integers.

Answer 6

For this particular problem, we might need the information for the cube roots of unity. The cube roots of unity are those values which when cubed give us 1 i.e. the solution of the equation: $$x^3=1$$ The values are : $x=1, \space -\frac{1}{2}+i\frac{\sqrt3}{2} \space -\frac{1}{2}-i\frac{\sqrt3}{2}$. Usually they are denoted as $x=1, \space \omega, \space \omega^2$. The good thing about this is if you take one complex root the square of it gives the other complex root.

Using this let us say you want to find the value of $(\frac{1}{2}+i\frac{\sqrt3}{2})^n$. We can write: $$(-\frac{1}{2}-i\frac{\sqrt3}{2})^n=\omega ^n$$ Therefore, $$(\frac{1}{2}+i\frac{\sqrt3}{2})^n=(-1)^n\omega ^n$$

Now the good thing about the powers of this complex number is that there will be a periodicity after some terms. Now we will put $n$

$n=0$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^0=1$$ $n=1$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^1=(-1)^1\omega ^1=-\omega$$ $n=2$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^2=(-1)^2\omega ^2=\omega^2$$ $n=3$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^3=(-1)^3\omega ^3=-\omega^3 = -1\space \text{(Cube root of unity)}$$

$n=4$

$$(\frac{1}{2}+i\frac{\sqrt3}{2})^4=(-1)^4\omega ^4=\omega^4=\omega * \omega^3 = \omega$$ $n=5$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^5=(-1)^5\omega ^5=-\omega ^5 = - \omega ^3 * \omega^2 = -\omega ^2$$ $n=6$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^6=(-1)^6\omega ^6=\omega ^ 6 = (\omega ^3)^2=1$$ $n=7$ $$(\frac{1}{2}+i\frac{\sqrt3}{2})^7=(-1)^7\omega ^7= - \omega^6 * \omega = -\omega$$ From this pattern you can see that the periodicity of the above function that you want to calculate is 6 i.e. after every $n+6$ the value is the same. So you can generalize the function. Once done put back the value of $\omega$ to get the cartesian form. Hope this helps...