Fools101
Imaginary Numbers
Simplify
i 9
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Fools101
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Please help!
pottersheep
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That is simplified isnt it?
Fools101
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oh also it i^9
pottersheep
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ah. ok then so
1^2 = -1 right
Blank123
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simplified version has to be i
Fools101
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Aww idk this kind of confusing me
pottersheep
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(i^2)^4 . i
(-1)^4 . i
1i
pottersheep
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(i^2)^4 . i
(-1)^4 . i
1i
Blank123
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any variable equals 1 automatically
pottersheep
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Do you get it?
pottersheep
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Basically take out an i^2 first so that you work with -1.
I.e i^4 = (i^2)^2
so now you have (-1)^2
=
1.
Do you understand up to there?
Fools101
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KIND of so it 1? a little ...
Blank123
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if there is no value to a variable it becomes a 1.
Fools101
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then the i Crossout ?
myininaya
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\[i^9=i^{4 \cdot 2 +1}=i^{4 \cdot 2} \cdot i^1=(i^4)^2 \cdot i=(1)^2 \cdot i\]
\[=1 \cdot i=i\]
anonymous
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pattern is
\[i^0=1\]
\[i^1=i\]
\[i^2=-1\]
\[i^3=-i\]
\[i^4=1\] and so on. so your real job is to take the integer remainder when you divide the power by 4. for example
\[i^{103}=i^3=-i\]
Fools101
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ohhhh i see !!! kk thank u that make sense !!!
anonymous
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yw