+0  
 
0
473
12
avatar+2499 

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
avatar+118687 
+2

i have just looked at the first one.

 

Hint:

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

 Nov 25, 2021
 #2
avatar+2499 
0

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
avatar+118687 
+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
avatar+2499 
0

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

Solveit  Nov 25, 2021
 #5
avatar+2499 
0

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
avatar+118687 
+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
avatar+2499 
0

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

Solveit  Nov 25, 2021
edited by Solveit  Nov 25, 2021
 #12
avatar+118687 
0

You got it !!  

Melody  Nov 25, 2021
 #7
avatar+118687 
0

I will look at the send one a little later.   smiley

 Nov 25, 2021
 #8
avatar+118687 
+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
avatar+33661 
+3

These are usually best done by using polar coordinates. e.g.

 

 Nov 25, 2021
 #10
avatar+118687 
+1

Thanks Alan,

 

I always forget how this is done   blush

Melody  Nov 25, 2021

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