Question

Can someone Please help me with imaginary Numbers? i beg you..

Answer 1

Alright, so what was your solution to \((-i)^3 (i^2)^3\)?

Answer 2

-i

Answer 3

Looks good.

Answer 4

yay :) can u give me a fly hard one

Answer 5

Let's combine a few things together.\[{i^2\cdot (-i)^5 \cdot \sqrt{-9}}\]

Answer 6

3i ...ugh probably wrong

Answer 7

You're close. I think you missed a power of \(i\). Check one more time. To help you along a bit, you can rewrite part of it as \[i^2(-i)^5=(-1)^5i^7\]

Answer 8

ok i will figure it out. i will be back in 10 mins ok? baby is crying downstairs

Answer 9

sure thing.

Answer 10

wow so sorry i am late

Answer 11

no worries.

Answer 12

3i x 1 sqrt -1

Answer 13

So \(3i\cdot \sqrt{-1}\)
Looks good. Can you simplify it a bit more?

Answer 14

so 3i square

Answer 15

how do you do square root (-2/7)

Answer 16

\(3i\cdot\sqrt{-1}=3i\cdot i=3i^2=-3\)

\[\sqrt{-2\over7}={\sqrt{-2} \over \sqrt{7}}={i \sqrt2 \over \sqrt7}\]

Answer 17

teacher said the answer is 5i square root (14) / 7

Answer 18

but then again..teacher = idiot

Answer 19

I'm a little confused about where the 5 came from, but generally people make sure to get radicals out of the denominator. So multiply the expression by \(\sqrt7/\sqrt7\)\[{i \sqrt2 \over \sqrt7}\cdot{\sqrt7 \over \sqrt7}={i \sqrt{2\cdot7} \over 7}={i\sqrt{14} \over 7}\]

Answer 20

oh okay! wow good think you told me that she takes of a lot of points if the radical is in the denominator ij use remembered lol

Answer 21

for i ^ 42 and i^91 what is a faster way to do that besides like writing out all of it?

Answer 22

Here's a neat little pattern for powers of \(i\). \[i^1=i\]\[i^2=-1\]\[^3=-i\]\[i^4=1\]\[i^5=i\]\[i^6=-1\]\[\vdots\]This pattern continues forever and ever. So if the power of \(i\) is divisible by 4, it becomes 1. If the power is even, but not divisible by 4, it's -1.

If the power is odd, and if you subtract 1 it's a multiple of 4, it becomes \(i\), and if it's odd, if you add 1 it's a multiple of 4, it becomes \(-i\).

Answer 23

Can you extend this pattern to \(i^{42}\) and \(i^{91}\)?

Answer 24

hmm im kind of confused can you give me an example of each of those four rules?

Answer 25

\[i^{25}=i\]Since 25-1=24 which is a multiple of 4. \[i^{66}=-1\]Since 66 is even, but not divisible by 4.\[i^{55}=-i\]Since 55+1=56 is divisible by 4.\[i^{84}=1\]Since 84 is divisible by 4.

Answer 26

wow perfect! just what i needed. could you give me a few practice ones?

Answer 27

How about \[i^{17}\]or\[i^{70}\]

Answer 28

i and -1

Answer 29

Perfect. What did you get for \(i^{42}\) and \(i^{91}\)?

Answer 30

-1 -i

Answer 31

Looks great.

Answer 32

woot! im gona do a few more ill chat you on here if i have more questions

Answer 33

sounds good.

Answer 34

\[(8-\sqrt{-11}) (8+ \sqrt{-11}) \]

Answer 35

This would be a place for FOIL. Could you give it a try before I show you how?

Answer 36

idk what is square root -11 squared

Answer 37

oh ok nvm 75

Answer 38

(4+3i)(2-5i)(4-3i)

Answer 39

Hold on. Go back to what we did earlier. \(\sqrt{-11}^2=-11\). Oh wait. You're evaluating the entire thing. Looks good.

Answer 40

For, that next one. Rewrite it as (4+3i)(4-3i)(2-5i) and FOIL (4+3i)(4-3i) first.

Answer 41

answer = 50 - 125 i ?

Answer 42

Looks perfect.

Answer 43

yay!!!!!

Answer 44

for this one it says find values for x and y in which each equation is true

Answer 45

3x-5yi = 15 - 20 i

Answer 46

Assuming x, y are real numbers and not complex, this means that you need to find an x, y such that \[3x=15\]\[-5y=-20\]
I've got to leave for a while now. Great job so far. It's a lot nicer for me when the people are actually trying to learn.

Answer 47

okay i will talk to you another time. thank yo so much for all your help i really appreciate it!