Let z 1 , z 2 , and z 3 be three complex numbers such that
∣ z 1 ∣ = ∣ z 2 ∣ = ∣ z 3 ∣ = 1
and
z 1 , z 2 , z 3 ∑ z 2 z 3 z 1 2 = − 1
Find the sum of all possible values of ∣ z 1 + z 2 + z 3 ∣ .
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Hi Aaghaz, how are you... Can you explain how did you write ∣ z 3 ∣ = ∣ z ∣ 3 while taking modulus?
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Heyyy!!! Long time no see dude!!! Kyaa chal rha hai aajkal??? Iss saal Prmo diya thaaa?? And what abt other Olympiads?? Syllabus kitna ho gyaa??
Anyways, regarding your doubt, as we know that for any two complex numbers, ∣ z 1 ⋅ z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ . And thus, you can easily extend this identity to any number of complex numbers to get ∣ ∣ ∣ ∣ ∣ j = 1 ∏ n z j ∣ ∣ ∣ ∣ ∣ = j = 1 ∏ n ∣ z j ∣ .
So by taking every z j = z we get ∣ z n ∣ = ∣ z ∣ n for any integer n and any complex number z .
Hope it helps!!
Arrey haan... Kaafi silly doubt tha.. Thanks bro.
Dekh, syllabus yahaan khatam ho gaya and revision started... And Maine prmo nahi Diya lekin baaki olympiads Dene Hain like KVPY, all NSEs...
What about you? Tumhara Kya chal Raha hai...
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Main to bas chillax kar rhaa hun......basically I don't like the notion of coaching vagarah, so I just go to FIITJEE and only give tests (usually) ......baaki sab I study at home........so mere according to I have done my syllabus......just Chemistry thodi problem de rhi hai (then again, kisko NAHI deti voh problem...!!).....I am also going to give all the olympiads.....KVPY,NSEP,NSEA, and NSEC......!! Slack jabse band hua uske baad tere se and pranav se baat hi nhi hui!!!
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Classes attend nahi karta? Agar doubts hote Hain fir kaise manage karta hai...
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@Vilakshan Gupta – Ummm...dude usually hote nahi hai so isliye.....anyways, sunn, tu Google Hangouts pe hai kya??? We can talk over there.....
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@Aaghaz Mahajan – Google Hangouts par to nahi Hun...
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@Vilakshan Gupta – lmao.....aaja fir......website pe jaa ke.....cuz teri email to hain and maine invite bhi bhejdiya
Also try my latest question...
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Lmao just solved it...kaafi easy thaa...XD
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Method Kya tha
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@Vilakshan Gupta – Come on hangouts.....just sent u an invite
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@Aaghaz Mahajan – I'm using phone... Aur app hai nahi... Aur website khulegi nahi.... Isliye chorh... And for downloading the app... I need space which I don't have..!! Lol
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@Vilakshan Gupta – WOW!!!!! Damnnn!!! XD!! Accha no problem then we can talk here only.....Main abhi solution likh rhaa hun teri problem kaa......wait a minute or so.....btw, tab tak tu meri physics waali problem check kar......!!! Yaa fir calculus waali bhi hain do......try those!
@Vilakshan Gupta – There dude........uploaded my solution.....
@Vilakshan Gupta – Solution samajh aa gyaa??
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@Aaghaz Mahajan – I knew this solution...main doosra solution jaanna chahta tha.. that's why I posted...but good..did u think it yourself when doing for the first time...I couldn't.....
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@Vilakshan Gupta – Main hangouts par aa gaya...
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Note that the summation can be manipulated and written as
z 1 3 + z 2 3 + z 3 3 − 3 z 1 z 2 z 3 = − 4 z 1 z 2 z 3
Now, factorising the left side, we get
( z 1 + z 2 + z 3 ) ( ( z 1 + z 2 + z 3 ) 2 − 3 ( z 1 , z 2 , z 3 ∑ z 1 z 2 ) ) = − 4 z 1 z 2 z 3
Now, because z ⋅ ( ∼ z ) = ∣ z ∣ 2 for any complex number z we have
( z 1 + z 2 + z 3 ) ( ( z 1 + z 2 + z 3 ) 2 − 3 z 1 z 2 z 3 ( ∼ ( z 1 + z 2 + z 3 ) ) ) = − 4 z 1 z 2 z 3
Now, putting z = z 1 + z 2 + z 3 we get
z 3 = ( 3 ∣ z ∣ 2 − 4 ) z 1 z 2 z 3
Finally, taking modulus on both sides gives us
∣ z ∣ 3 = ∣ ∣ ∣ 3 ∣ z ∣ 2 − 4 ∣ ∣ ∣
Solving this equation gives us ∣ z ∣ = − 2 , 1 , 2 . But since ∣ z ∣ ≥ 0 , we have
∣ z 1 + z 2 + z 3 ∣ = 1 , 2
Note:
Here ∼ z denotes the conjugate of the complex number z