+0

# how to calculate something to the 100th degree?

+1
266
12

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
+2

i have just looked at the first one.

Hint:

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

Nov 25, 2021
#2
+1

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
+2

$$(-\sqrt2+i\sqrt2)^{100}=((-\sqrt2+i\sqrt2)^{2})^{50}\\~\\ (-\sqrt2+i\sqrt2)^{2}=2-2-4i = -4i$$

.
Nov 25, 2021
edited by Melody  Nov 25, 2021
#4
+2

for the second one, I cannot divide 35 by 2 or is it what I should do?

Solveit  Nov 25, 2021
#5
+2

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
+2

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
+1

Okay, I got it. it is -1, therefore $$-(4)^{50}$$

Solveit  Nov 25, 2021
edited by Solveit  Nov 25, 2021
#12
0

You got it !!

Melody  Nov 25, 2021
#7
0

I will look at the send one a little later. Nov 25, 2021
#8
+1

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
+3

These are usually best done by using polar coordinates. e.g. Nov 25, 2021
#10
+1

Thanks Alan,

I always forget how this is done Melody  Nov 25, 2021