# solving complex numbers with powers algebraically

Find algebraically the value of :$\left(2^{0.5} + 6^{0.5} - \left( 2^{0.5} - 6^{0.5} \right)i \right)^4$

Below are my works

I try to simplify inside. but i found that i can't add $2^{0.5}$ and $6^{0.5}$ together.

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2^0.5 is square root of 2 btw – carry Mar 17 '14 at 5:31
the answer sheet says the answer is 256i ... so the answer is wrong ? – carry Mar 17 '14 at 5:53
The real part of the expression in the question is not zero (on first glance). – Lee Yiyuan Mar 17 '14 at 6:06
WA apparently agrees with my answer: $\;128(1+\sqrt3\,i)\;$ ...wolframalpha.com/input/… – DonAntonio Mar 17 '14 at 6:34
Oh, and btw: yes, that answer sheet's answer is wrong. – DonAntonio Mar 17 '14 at 6:38

Simplify by manipulating the inner expression:

\begin{align}\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})i &= (\sqrt{2} + \sqrt6i) + (\sqrt{6} - \sqrt2i)\\ &= (\sqrt{2} + \sqrt6i) - (\sqrt{2} + \sqrt6i)i \\&= (\sqrt{2} + \sqrt6i)(1 - i)\end{align}

Now, let \begin{align}z &= (\sqrt{2} + \sqrt{6} - (\sqrt{2} - \sqrt{6})i)^4 \\ &= (\sqrt{2} + \sqrt6i)^4(1 - i)^4\end{align}

Then, \begin{align}|z| &= |\sqrt{2} + \sqrt6i|^4\cdot|1 - i|^4 \\ &= (\sqrt{2 + 6})^4 \cdot (\sqrt{1 +1})^4 \\ &= 256\end{align}

On the other hand, \begin{align}\arg{z} &= 4\arg(\sqrt{2} + \sqrt6i) + 4 \arg(1 - i) \\ &= 4 \tan^{-1}\frac{\sqrt{6}}{\sqrt{2}} -4 \tan^{-1}1 \\ &= \frac{\pi}{3}\end{align}

Hence

$$z = 256e^{i\frac{\pi}{3}}$$

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Hints:

$$\sqrt2\pm\sqrt6=\sqrt2(1\pm\sqrt3)\implies \left(\sqrt2+\sqrt 6-(\sqrt2-\sqrt6)i\right)^4=$$

$$=4\left(1+\sqrt3-(1-\sqrt3)i\right)^4=4\left[(1+\sqrt3)^2-(1-\sqrt3)^2-2(1+\sqrt3)(1-\sqrt3)i\right]^2=$$

$$=4\left[4\sqrt3+4i\right]^2=64(\sqrt3+i)^2=64(2+2\sqrt3i)=\ldots$$

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Set $\displaystyle\sqrt2+\sqrt6=R\cos\phi, \sqrt6-\sqrt2=R\sin\phi$ where real $R\ge0$

Squaring and adding we get $\displaystyle R^2=2(2+6)=16\implies R=4$

On division, $\displaystyle\tan\phi=\frac{R\sin\phi}{R\cos\phi}=\frac{\sqrt6-\sqrt2}{\sqrt6+\sqrt2}=\frac{\sqrt3-1}{\sqrt3+1}=2-\sqrt3$ which is $\tan15^\circ$ (Find here)

From the dentition of atan2, $\displaystyle\phi=15^\circ$

So, we have $\sqrt2+\sqrt6+(\sqrt6-\sqrt2)i=4(\cos15^\circ+i\sin15^\circ)$

Now apply de Moivre's Identity

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I'll put in that if one didn't have the exact value for $\tan 15º \$ handy, one could use a right triangle (or trig identities) to establish that $\ \sin \theta = \frac{2 - \sqrt{3}}{\sqrt{8 - 4 \sqrt{3}}} \$ and $\ \cos \theta = \frac{1}{\sqrt{8 - 4 \sqrt{3}}} \ .$ As horrible as this looks at the moment, the "double-angle formula" for sine produces $$\sin 2 \theta \ = \ \frac{2 \cdot \ (2 - \sqrt{3}) \cdot 1}{8 - 4 \sqrt{3}} \ = \ \frac{4 - 2\sqrt{3}}{8 - 4 \sqrt{3}} \ = \ \frac{1}{2} \ .$$ Now we know where we are, and can get the angles we need for the fourth-power result. – RecklessReckoner Mar 17 '14 at 7:25