MedVision ad

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

amdspotter

Member
Joined
Dec 9, 2020
Messages
126
Gender
Male
HSC
2022
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

Well-Known Member
Joined
Oct 15, 2021
Messages
562
Gender
Male
HSC
2018
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

Member
Joined
Dec 9, 2020
Messages
126
Gender
Male
HSC
2022
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

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,390
Gender
Male
HSC
2006
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

Well-Known Member
Joined
Oct 7, 2021
Messages
784
Gender
Male
HSC
2022
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

Moderator
Moderator
Joined
Mar 11, 2004
Messages
2,642
Gender
Male
HSC
N/A
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.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top