It's complex
The maximum value of ∣ z ∣ when z satisfies the condition ∣ z + z 2 ∣ = 2
Note: 3 = 1 . 7 1
The answer is 2.71.
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2 solutions
Let ∣ z ∣ = r then by triangle inequality:- ( r − r 2 ) ≤ ∣ z + z 2 ∣
So
( r − r 2 ) ≤ ∣ z + z 2 ∣ = 2 Hence
( r − r 2 ) ≤ 2
Solving for equality case we get
r = 1 + 3 a n d r = 1 − 3
Hence maximum value is 1 + 3
2 . 7 1
Good solution
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FYI: Equations tend to appear better if you place all of them in 1 latex bracket, instead of leaving some stuff outside. I have edited your question for reference.
You still need to show that r = 1 + 3 can indeed be achieved. It is not immediately obvious why this is so.
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r − r 2 ≤ 2
r 2 − 2 ≤ 2 r
Taking the case of equality
r 2 − 2 r − 2 = 0
r=[ 2 ± 1 2 ]/ 2
So
r = 1 ± 3
r must be positive
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Note, you still have not shown that there is a complex number z such that ∣ z ∣ = 1 + 3 and ∣ z + z 2 ∣ = 2 .
That is what is meant by showing that the equality case can indeed be achieved.
Nicely done .
Triangle inequality theorem is | z1 | + | z2 | <= | z1 + z2 |
Then how can that be [ r -( r/2 ) ] < = | z + (2/z ) | ?
i squared both sides and got r^2+1/r^2 + 2Re(2z/(z bar) )=4 . using euler's thm.t is argument. r^2+1/r^2 + 4cos(2t)=4 diff. r wrt t and dr/dt=0.Then u get 2rdr/dt - 1/r^3 dr/dt-8sin2t=0.Put dr/dt=0 and get t=npi/2 .Just use quadratc formula and express r in terms of sint bcos cos2t=2cos^2t + 1 .U will get ANS