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# how to calculate something to the 100th degree?

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I saw a calculation by putting the opposite like: (-sqrt(2)-isqrt(2))^2, but I am not sure if I should do this one the same way.

Nov 25, 2021

#1
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i have just looked at the first one.

Hint:

$$(-\sqrt2+i\sqrt2)^{100}=((-\sqrt2+i\sqrt2)^{2})^{50}$$

Nov 25, 2021
#2
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yep but opening braces goes like this:

$$(2-4*i+2)^{50}$$

I have no idea how to proceed from here and I don't know if I opened the braces correctly

Solveit  Nov 25, 2021
#3
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$$(-\sqrt2+i\sqrt2)^{100}=((-\sqrt2+i\sqrt2)^{2})^{50}\\~\\ (-\sqrt2+i\sqrt2)^{2}=2-2-4i = -4i$$

.

Nov 25, 2021
edited by Melody  Nov 25, 2021
#4
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for the second one, I cannot divide 35 by 2 or is it what I should do?

Solveit  Nov 25, 2021
#5
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okay so the answer is 4^{50}, I understood the first one.

Solveit  Nov 25, 2021
edited by Solveit  Nov 25, 2021
edited by Solveit  Nov 25, 2021
#6
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That is not quite right.  it is (-4i)^50 = (-4)^50 * i^50 = 4^50 * i^50

You need to work out what i^50

once again you do it with a pattern

hint:

i^1=

i^2=

i^3=

i^4=

i^5=

what is the pattern?

i^ 50 = ?

Melody  Nov 25, 2021
#11
+2578
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Okay, I got it. it is -1, therefore $$-(4)^{50}$$

Solveit  Nov 25, 2021
edited by Solveit  Nov 25, 2021
#12
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You got it !!

Melody  Nov 25, 2021
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I will look at the send one a little later.

Nov 25, 2021
#8
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I mucked around quite a lot with the second one before I got it.  (Maybe there are some shortcut techniques)

anyway:

It should help you a lot if you work out the value of        $$(1-\sqrt3i)^3$$

Nov 25, 2021
#9
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These are usually best done by using polar coordinates. e.g.

Nov 25, 2021
#10
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Thanks Alan,

I always forget how this is done

Melody  Nov 25, 2021