Question

Please help! :)

Answer 1

kay

Answer 2

\(i^2=-1\)

Answer 3

-6-17i

Answer 4

i think

Answer 5

@bbcream14 not correct

Answer 6

okay then give me a sec

Answer 7

So.. i^10 = -1 ?

Answer 8

yes. \[i^{10} = i^2*i^2*i^2*i^2*i^2 = (-1)(-1)(-1)(-1)(-1) = -1\]

Answer 9

Okay now what?

Answer 10

hint: compute [(1 - i)^2 ]^5

Answer 11

Yeah but these are the answers..
-32i
-32 + 32i
32
32 -32i

Answer 12

-.-

those are the answers AFTER you do some simplification

Answer 13

i'll give you the next hint:
(1-i)^2 = 1 - 2i + i^2, and i^2 = -1

Answer 14

Oh okay.

Answer 15

i dont get where you got the 1 - 2i + 1^2 ...

Answer 16

because you foil (1-i)^2 you'll get 1 - 2i + i^2
since i^2 = -1, you have 1 - 2i - 1 = -2i

and so (-2i)^5 = (-2)^5 * i^5 = -32i

Answer 17

Oh duh

Answer 18

Or you could expand \[(1-i)^{10} = i^{10}-10 i^9+45 i^8-120 i^7+210 i^6-252 i^5+210 i^4-120 i^3\]\[\qquad+45 i^2-10 i+1\] and simplify :-)

Answer 19

Still trying to understand the rest haha

Answer 20

This is definitely a problem where thinking a bit first will save you a lot of effort!

Answer 21

Yes! haha can you help me with another one? i still dont understand it fully

Answer 22

(-√3 + i)^6

Answer 23

Im still trying to understand this though.. (1−i)10=i10−10i9+45i8−120i7+210i6−252i5+210i4−120i3
+45i2−10i+1

Answer 24

Can you help me with this one? (-√3 + i)^6
]

Answer 25

If you multiply out \[(1-i)^{10}=(1-i)(1-i)(1-i)(1-i)(1-i)(1-i)(1-i)(1-i)(1-i)(1-i)\]that's what you'll get....but then those various powers of \(i\) simplify to
\[-1-(-10i)+45 +120i \text{ etc.}\]

Answer 26

Ohhh okay i get it :)