Complex + Radical Evaluation

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#1 h Mar 17, 2025, 1:08 PM

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Evaluate: (without calculators)
$(\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$

This post has been edited 1 time. Last edited by Saucepan_man02, Mar 17, 2025, 1:08 PM
Reason: EDIT

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#2 h Mar 17, 2025, 1:30 PM

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I think you just find the magnitude of each one, then evaluate the angle, and then double it by 5 then evaluate the trig of it to find it

too lazy to evaluate it

This post has been edited 1 time. Last edited by MathRook7817, Mar 17, 2025, 1:31 PM

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#3 h Apr 1, 2025, 10:46 AM

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Have this a nice solution?

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#5 h Apr 1, 2025, 4:45 PM • 1 Y

Y by Mathzeus1024

Let $\sqrt{6 - 2 \sqrt{5}} = x$ and $i \sqrt{2 \sqrt{5} + 10} = y$ .

Then we have $(x+y)^5 + (x-y)^5$ . Multiplying out and combining like terms, we get $2x^5+20x^3y^2+10xy^4$ .

Factoring out $2x$ gives us $2x(x^4+10x^2y^2+5y^4)$ .

Substitute $\sqrt{6 - 2 \sqrt{5}}$ and $i \sqrt{2 \sqrt{5} + 10}$ back in for $x$ and $y$ .

Substituting for $x$ and $y$ gives us
$2( \sqrt{6 - 2 \sqrt{5}})[(\sqrt{6 - 2 \sqrt{5}})^4 + 10( \sqrt{6 - 2 \sqrt{5}})^2(i \sqrt{2 \sqrt{5} + 10})^2 + 5(i \sqrt{2 \sqrt{5} + 10})^4]$

Multiplying out, we have
$2( \sqrt{6 - 2 \sqrt{5}})[(56 - 24 \sqrt{5}) + 10(-40 + 8 \sqrt{8}) + 5(120 + 40 \sqrt{5})]$
$= 2( \sqrt{6 - 2 \sqrt{5}})(256 + 256 \sqrt{5})$

We can write $6 - 2 \sqrt{5}$ as the square of some $a + b$ , so we have $6 - 2 \sqrt{5} = a^2 + 2ab + b^2$ .

Since $a^2$ and $b^2$ are most likely integers, then we have $-2 \sqrt{5} = 2ab$ . $ab$ is then equal to $- \sqrt{5}$ , giving us $1$ and $- \sqrt{5}$ or $-1$ and $\sqrt{5}$ . (It really doesn't matter what order they're in)

So $\sqrt{6 - 2 \sqrt{5}}$ equals either $1 - \sqrt{5}$ or $\sqrt{5} -1$ . I would assume that we use the positive root, but if not, we can just negate our final answer to get the other possible answer.

The positive root of $\sqrt{6 - 2 \sqrt{5}}$ is $\sqrt{5} -1$ , so we have
$2( \sqrt{5} -1)(256 + 256 \sqrt{5})$
$= 2(256)( \sqrt{5} -1)( \sqrt{5} +1)$
$= (512)(4)$
$= 2048$

The other answer would be $-2048$ if you use the negative square root, since it's just the negation of the positive root we used.

This post has been edited 1 time. Last edited by SmartHusky, Apr 1, 2025, 10:06 PM
Reason: Finished sol after school

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#6 h Apr 2, 2025, 11:47 AM

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Let $x=\sqrt{6-2\sqrt{5}}, y=\sqrt{2\sqrt{5}+10} \Rightarrow \sqrt{x^2+y^2} = \sqrt{(6-2\sqrt{5})+(2\sqrt{5}+10)} = 4$ . Using the Eulerian form $z=x+yi = \sqrt{x^2+y^2}e^{i \cdot \arctan(y/x)}$ , we can rewrite the above expression according to:

$(\sqrt{x^2+y^2}e^{i \cdot \arctan(y/x)})^5+(\sqrt{x^2+y^2}e^{-i \cdot \arctan(y/x)})^5$ ;

or $4^{5}[e^{i \cdot 5\arctan(y/x)})^5+e^{-i \cdot 5\arctan(y/x)}]$ ;

or $2^{10} \cdot 2\cos\left[5\arctan \left(\frac{y}{x}\right)\right]$ (i).

Using the identities:

$\cos(5t) = 5\cos(t)-20\cos^{3}(t)+16\cos^{5}(t)$ (ii);

$t=\arctan\left(\frac{y}{x}\right) = \arccos\left(\frac{x}{\sqrt{x^2+y^2}}\right) = \arccos\left(\frac{\sqrt{6-2\sqrt{5}}}{4}\right) = \arccos\left(\frac{1}{4}\sqrt{(\sqrt{5}-1)^{2}}\right) = \arccos\left(\frac{\sqrt{5}-1}{4}\right)$ (iii)

in (i) yields:

$2^{11}\left[5\left(\frac{\sqrt{5}-1}{4}\right) -20\left(\frac{\sqrt{5}-1}{4}\right)^{3}+16\left(\frac{\sqrt{5}-1}{4}\right)^{5}\right]$ ;

or $\frac{2^{11}}{2^{10}} \cdot [5(4^4)(\sqrt{5}-1)-20(4^2)(\sqrt{5}-1)^3+16(\sqrt{5}-1)^5]$ ;

or $2[1280(\sqrt{5}-1)-320(8\sqrt{5}-16)+16(80\sqrt{5}-176)]$ ;

or $2(1024) = \textcolor{red}{2048}$ .

This post has been edited 4 times. Last edited by Mathzeus1024, Apr 2, 2025, 11:56 AM

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