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Confusion ABT how to prove something is purely imaginary (1 Viewer)

amdspotter

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Need to find values of n for which (1+I)^n is purely real or imaginary. Did the real part but imag part always screws me up so yh wanted to know how to do it.
I did this so farIMG_20211202_224245.jpg
 

5uckerberg

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Answers say -2(2k+1)
Well does it matter I mean it does to an extent but at the end of the day they are still the same thing. Since cosine is an even function positive or negative should give us the same thing.
 

amdspotter

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Well does it matter I mean it does to an extent but at the end of the day they are still the same thing. Since cosine is an even function positive or negative should give us the same thing.
Wait can U explain the process through which U determined it's 2 + 4k
 

Trebla

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The key is to realise that what you are writing is the general term of an infinite sequence of numbers.

When you write , you actually are writing the general term for n = {...,-8,-4,0,4,8,...}. In fact, since k is any integer it doesn't matter whether we write n = 4k or n = -4k, they each generate the same sequence n = {...,-8,-4,0,4,8,...}.

This means that n = 4k and n = -4k are technically equivalent, so you could've written just n = 4k only (or n = -4k only) and that would suffice. You could even write it like n = 4(k+1) where k is an integer (i.e. replace k with k+1). This is still valid as it generates the same sequence of values n = {...,-8,-4,0,4,8,...}.

You can apply the same logic for the other part.

From your attempt on the purely imaginary part, it gives (which is correct) which represents the sequence n = {...,-10,-6,-2,2,6,10,...}.

Since k is any integer, you could replace k with k+1 and rewrite n = 4k-2 as n = 4(k+1)-2 = 4k+2. This suggests that n = 4k-2 and n = 4k+2 are in fact equivalent as they each generate the same sequence n = {...,-10,-6,-2,2,6,10,...} so you could've just written n = 4k+2 only (or n = 4k-2 only) and that would suffice.

We also could have replaced k with -k (given k is any integer), so you could've written n = -4k+2 only (or n = -4k-2 only). Either way, it doesn't matter which way you index it, they all represent the same sequence of numbers n = {...,-10,-6,-2,2,6,10,...}.
 

Run hard@thehsc

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I observed that for purely imaginary number cases, it is generally something times (k + 1) set of values - is this the case for most problems?
 

CM_Tutor

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I observed that for purely imaginary number cases, it is generally something times (k + 1) set of values - is this the case for most problems?
I wouldn't try and remember a most-cases solution, derive it from real part equals zero on a case-by-case basis.
 

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