# Find the 2016th power of a complex number

Calculate (−1+i√31+i)2016\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}.

Here is what I did so far:

I’m trying to transform zz into its trigonometric form, so I can use De Moivre’s formula for calculating z2016z^{2016}.

Let z=−1+i√31+iz = \frac{-1 + i\sqrt 3}{1 + i}. This can be rewritten as √3−12+i√3+12\frac{\sqrt 3 – 1}{2} + i\frac{\sqrt 3 + 1}{2}.

z=r(cosϕ+isinϕ)z = r(\cos \phi + i \sin \phi)

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

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

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Can you double check that argument again?
– imranfat
2 days ago

I think it is ϕ=arctan(√3+2)\phi = \arctan( {\sqrt 3 + 2})
– guestDiego
2 days ago

44

Wow, these math problems are getting harder! When I was in school we only had to compute the 1975th power of this complex number 🙂
– Dave L. Renfro
2 days ago

2

ϕ=arctan√3+1√3−1\phi = \arctan \frac {\sqrt 3 +1}{\sqrt3 -1}
– Doug M
2 days ago

2

@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. 🙂
– Deepak
yesterday

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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(−√3)\arctan(-\sqrt{3}) as well as arctan1\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=√2r=\sqrt{2} and θ=45آ°\theta=45آ°, so to the power 2016 is 2^{1008}(\cos(2016(45آ°))+i\sin(2016(45آ°)))2^{1008}(\cos(2016(45آ°))+i\sin(2016(45آ°))) which is 2^{1008}(1+0i)2^{1008}(1+0i). Can you do the numerator?

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

\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 }\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 }

Note that

z=
\frac{-1 + i\sqrt 3}{1 + i} =
\frac{-1 + i\sqrt 3}{2}
\frac{2}{1 + i}
=\sqrt2 \alpha \beta

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},
\beta = \frac{\sqrt2}{1 + i}

\alpha=\frac{-1 + i\sqrt 3}{2},
\beta = \frac{\sqrt2}{1 + i}

Note that \alpha^3 = 1 = \beta^8 \alpha^3 = 1 = \beta^8 .

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

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} \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}} \frac{\pm 1 \pm \mathbf{i}}{\sqrt{2}}
Sometimes people like to write \frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2} rather than \frac{1}{\sqrt{2}}\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} \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} \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\zeta_n to denote the principal nn-th root of unity.

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

Sorry for the noise
– lhf
yesterday

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

Lets simplify \frac{-1 + i\sqrt 3}{1 + i}\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)\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)\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}\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} )(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = (\cos \frac {-2016\pi}{4} +\sin \frac {-2016\pi}{4} )

88 divides 20162016

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

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

66 divides 20162016, too.

2^{1008}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})\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.

Note that

\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} \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} \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}\left (\sqrt{2}\right)^{2016} \left((cos(840\pi) + i \cdot sin(840\pi)) \right)\implies 2^{1008}

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).
– Hurkyl
yesterday