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Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.

Here is what I did so far:

I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.

Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 - 1}{2} + i\frac{\sqrt 3 + 1}{2}$.

$$z = r(\cos \phi + i \sin \phi)$$

$$r = |z| = \sqrt 2$$ $$\phi = \arctan {\sqrt 3 + 1}$$

Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?

Thank you in advance!

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  • $\begingroup$ Can you double check that argument again? $\endgroup$
    – imranfat
    Oct 21, 2016 at 17:53
  • $\begingroup$ I think it is $\phi = \arctan( {\sqrt 3 + 2})$ $\endgroup$
    – guestDiego
    Oct 21, 2016 at 17:54
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    $\begingroup$ Wow, these math problems are getting harder! When I was in school we only had to compute the 1975th power of this complex number :) $\endgroup$ Oct 21, 2016 at 18:22
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    $\begingroup$ $\phi = \arctan \frac {\sqrt 3 +1}{\sqrt3 -1}$ $\endgroup$
    – Doug M
    Oct 21, 2016 at 18:32
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    $\begingroup$ @DaveL.Renfro +1 for making me LOL, and I wish I could give you another ten upvotes for not being afraid to reveal your vintage. :) $\endgroup$
    – Deepak
    Oct 22, 2016 at 10:22

7 Answers 7

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Here is another approach: Why don't you "distribute" that exponent on the numerator and denominator? Then raise both numerator and denominator to the power 2016. The thing is that both $\arctan(-\sqrt{3})$ as well as $\arctan 1$ are well known angles. From there you can apply your DeMoivre. Once you have those new numerators and denominators, you can simply divide. I will do the denominator for you: $r=\sqrt{2}$ and $\theta=45°$, so to the power 2016 is $2^{1008}(\cos(2016(45°))+i\sin(2016(45°)))$ which is $2^{1008}(1+0i)$. Can you do the numerator?

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$$\dfrac{\sqrt3-1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4-\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi6+\dfrac\pi4\right)$$

$$\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\sin\dfrac\pi4+\sin\dfrac\pi6\cos\dfrac\pi4=\sin\left(\dfrac\pi6+\dfrac\pi4\right)$$


OR

$$-1+\sqrt3i=2\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right)=2e^{2i\pi/3}$$

$$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)=\sqrt2e^{i\pi/4}$$

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$$\left( \frac { -1+i\sqrt { 3 } }{ 1+i } \right) ^{ 2016 }=\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2\pi }{ 3 } \right) +i\sin { \left( \frac { 2\pi }{ 3 } \right) } } } \right) }^{ 1013 } }{ { \left( { \left( 1+i \right) }^{ 2 } \right) ^{ 1013 } } } =\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2016\pi }{ 3 } \right) +i\sin { \left( \frac { 2016\pi }{ 3 } \right) } } } \right) } }{ { 2 }^{ 1008 }{ i }^{ 1008 } } =\\ ={ 2 }^{ 1008 }{ e }^{ 2016i\pi /3 }={ 2 }^{ 1008 }$$

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Note that $$ z= \frac{-1 + i\sqrt 3}{1 + i} = \frac{-1 + i\sqrt 3}{2} \frac{2}{1 + i} =\sqrt2 \alpha \beta $$ where $$ \alpha=\frac{-1 + i\sqrt 3}{2}, \qquad \beta = \frac{\sqrt2}{1 + i} $$ Note that $\alpha^3 = 1 = \beta^8 $.

Therefore $z^{24}=2^{12}$ and so $z^{2016}=2^{1008}$.

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The main trick that makes this easy is to immediately recognize by sight parts of the third and eighth roots of unity.

The two primitive third roots of unity are $$ \frac{-1 \pm \mathbf{i} \sqrt{3}}{2} $$ and the four primitive eighth roots of unity are $$ \frac{\pm 1 \pm \mathbf{i}}{\sqrt{2}} $$ Sometimes people like to write $\frac{\sqrt{2}}{2}$ rather than $\frac{1}{\sqrt{2}}$.

And since they're similar, I'll mention the two primitive sixth roots of unity are $$ \frac{1 \pm \mathbf{i} \sqrt{3}}{2} $$

In each case, the root with smallest positive complex argument is the one where you take the positive signs.

With this in mind, we rewrite the base as

$$\frac{-1 + i\sqrt 3}{1 + i} = \frac{-1 + \mathbf{i} \sqrt{3}}{2} \frac{\sqrt{2}}{1 + \mathbf{i}} \sqrt{2} = \sqrt{2} \zeta_3 \zeta_8^{-1} $$

where I've used $\zeta_n$ to denote the principal $n$-th root of unity.

Now, the 2016-th power is easy to compute!

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  • $\begingroup$ Sorry for the noise $\endgroup$
    – lhf
    Oct 22, 2016 at 13:36
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$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$

Lets simplify $\frac{-1 + i\sqrt 3}{1 + i}$

$\frac {1}{1+i}(-1 + i\sqrt 3)\\ \frac {1-i}{2}(-1 + i\sqrt 3)\\ (1-i)(-\frac12 + i\frac{\sqrt 3}2)$

Convert to polar:

$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)$

Now we have a choice...we could raise to the 2016 power right now, or we could mulitiply those two complex numbers first then raise to the 2016 power.

$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} = \sqrt 2^{2016} (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016}(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)^{2016}$

Applying deMoivres theorem:

$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = (\cos \frac {-2016\pi}{4} +\sin \frac {-2016\pi}{4} )$

$8$ divides $2016$

$\frac {-2016\pi}{4} = 2n\pi$

$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = 1$

$6$ divides $2016$, too.

$2^{1008}$

alternatively:

$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3) = \sqrt 2 (\cos \frac {5\pi}{12} +\sin \frac {5\pi}{12})$

and we get to the same place.

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Note that

$$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} $$can be expressed as$$ \left( \sqrt{2}(cos(\frac{5\pi}{12}) + i \cdot sin(\frac{5\pi}{12})) \right)^{2016}$$

which gives

$$\left (\sqrt{2}\right)^{2016} \left((cos(840\pi) + i \cdot sin(840\pi)) \right)\implies 2^{1008}$$

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  • $\begingroup$ While true (presumably -- I didn't check), it sounds like the OP knows all this -- what the OP is missing is how to do that first step (and how to notice that you could do it). $\endgroup$
    – user14972
    Oct 22, 2016 at 9:41

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