Study the complex number $z = \left( \frac{\sqrt{3} - i}{1 + i} \right) ^ {12}$.

Question

I am given the complex number:

$$z = \bigg ( \dfrac{\sqrt{3} - i}{1 + i} \bigg ) ^ {12}$$

And I have to choose a true description of this number. Only one of the following descriptions is true:

A. $z = 2^6$

B. $\arg(z) = \pi$

C. $|z| = 2 ^ {12}$

D. $z = 64i$

E $\arg(z) = 2 \pi$

My problem is that I can't manipulate the number $z$ such that I can use DeMoivre's formula on $z$. This is as far as I got:

$$z = \bigg ( \dfrac{\sqrt{3} - i} {1 + i} \bigg ) ^ {12} = \bigg ( \dfrac{(\sqrt{3} - i) (1 - i)} {1 - i^2} \bigg ) ^ {12} = \bigg ( \dfrac{\sqrt{3} - \sqrt{3}i - i + i^2} {2} \bigg ) ^ {12} =$$

$$ = \bigg ( \dfrac{\sqrt{3} - 1 - (\sqrt{3} + 1)i} {2} \bigg ) ^ {12} = \bigg ( \dfrac{\sqrt{3} - 1} {2} + \dfrac{-(\sqrt{3} + 1)}{2} \bigg ) ^ {12}$$

And this is where I got stuck. I know that I need to get $z$ in a form that looks something like this:

$$z = (\cos \theta + i \sin \theta) ^ {12}$$

But I can't find an angle whose cosine equals $\dfrac{\sqrt{3} - 1}{2}$ and whose sine equals $\dfrac{-(\sqrt{3} + 1)}{2}$. So how can find the following:

$$\cos \hspace{.1cm} ? = \dfrac{\sqrt{3} - 1}{2}$$

$$\sin \hspace{.1cm} ? = \dfrac{-(\sqrt{3} + 1)}{2}$$

Answer 1

The key insight is to recognize roots of unity in the expression.

Let $a=1+i$, $b=\sqrt{3}-i$. Let $A=a/\sqrt 2$, $B=b/2$.

Then $A^4=-1$, $B^6=-1$, and so $$ z=\frac{b^{12}}{a^{12}} =\frac{2^{12} B^{12}}{2^6 A^{12}} =-2^{6} $$

Answer 2

Hint:

Write separately in exponential form $z_n=\sqrt 3-i$ and $z_d=1+i$. Remember the argument of a complex number is defined modulo $2\pi$.

Answer 3

My classic solution is to use the formula: $$\frac{a+bi}{c+di} = \frac{\left(c-di\right)\left(a+bi\right)}{\left(c-di\right)\left(c+di\right)} = \frac{\left(ac+bd\right)+\left(bc-ad\right)i}{c^2+d^2}$$

$$\frac{\sqrt{3}-i}{1+i}=\frac{\sqrt{3}-1+i\left(-1-\sqrt{3}\right)}{2}$$ and after I will calculate with a calculator $\left(\sqrt{3}-1+i\left(-1-\sqrt{3}\right)\right)^{12}$.

$$\left(\frac{\sqrt{3}-i}{1+i}\right)^{12}=\left(\frac{\sqrt{3}-1+i\left(-1-\sqrt{3}\right)}{2}\right)^{12}$$

Answer 4

Let $w=\dfrac{\sqrt3-i}{1+i}$.

The denominator squared should be recognizable; $w^2=\dfrac{2-2\sqrt3i}{2i}=-\sqrt3-i=2e^{-5i\pi/6},$

so $z=w^{12}=(2e^{-5i\pi/6})^6.$ Can you take it from here?

Answer 5

Rewrite what you obtained,

$$ \dfrac{\sqrt{3} - 1} {2} + \dfrac{-(\sqrt{3} + 1)}{2} i =\sqrt2 \left( \cos\theta + i \sin\theta\right)$$

where

$$\cos\theta = \dfrac{\sqrt{3} - 1} {2\sqrt2},\>\>\>\>\> \sin\theta = \dfrac{-\sqrt{3} - 1} {2\sqrt2}$$

Note

$$\sin2\theta=2\sin\theta\cos\theta= 2\cdot\frac{\sqrt{3} - 1 }{\sqrt2}\cdot\frac{-\sqrt{3} - 1 }{\sqrt2}=-\frac12$$

which leads to $2\theta=-\frac\pi6$, or $\theta = -\frac\pi{12}$. Then,

$$ z= \bigg ( \sqrt2 (\cos\frac\pi{12} - \sin\frac\pi{12} i)\bigg ) ^ {12} =2^6(\cos\pi - i\sin\pi )=2^6e^{i\pi}$$

Thus, $\arg(z) = \pi$

Answer 6

Let $w_1 = \sqrt{3} - i$ and $w_2 = 1 + i$. Then $$|w_1| = \sqrt {1^2 + (\sqrt{3})^2} = 2$$ and $$\arg w_1 = {5 \pi/3}$$ while $$|w_2| = \sqrt {1^2 + 1^2} = \sqrt {2}$$ and $$\arg w_2 = \pi /4$$

Then we have by DeMoivre's Theorem $$\left(\dfrac {w_1}{w_2}\right)^n \text { cis } \left(\arg n(w_1 - w_2\right))$$ or $$\dfrac {2^{12}}{{\sqrt {2}^{12}}} \text { cis } (12*(5\pi/3 - \pi/4)) \rightarrow 2^{(12-6)} \text { cis } (12*(17 \pi/12)) \rightarrow 2^6 \text { cis } \pi = -2^6$$ (where $\sqrt{2}^{12} = 2^{12/2} = 2^6$, and $17\pi = 8 \cdot 2\pi + \pi$ and $\text { cis } \pi = -1$)

The question asks which description of the number is true, and $\mathbf{B}$ is the correct answer ($\arg z = \pi$).