juantheron
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ok cool, yeh i used isos triangle too^ Yea pretty much
I think you can also use isosceles triangles?
There is something wrong with this question.
Draw a circle with three point A, B, C representing z1, z2, z3.What is the method of solving this?
|z1 - z2| doesn't equal |z1|-|z2|. I don't think something is wrong with the question.There is something wrong with this question.
Let z1=1, z2=1, z3=-1.
These satisfy the first requirement that they all have unit modulus.
So it satisfies the second requirement.
Now let z1=1, z2=i, z3=i.
Again the two requirements are met (you can check), but this time the answer is sqrt 2.
In fact I made a Geogebra construction, and you can get an infinite number of answers.
I think the question might be correct if you had asked for the MAXIMUM value of
I don't see where he has used this?|z1 - z2| doesn't equal |z1|-|z2|. I don't think something is wrong with the question.
|z1-z2|^2 = |1-1|^2I don't see where he has used this?
Nah everything he did was correct.|z1-z2|^2 = |1-1|^2
I think that is a nonsense regardless
Not sure what it is you don't get, but all I've done is replace z1 by 1 and z2 by 1. Simple substitution. If you think year 7 substitution is nonsense, well ... what can I say.|z1-z2|^2 = |1-1|^2
I think that is a nonsense regardless
Sorry, my bad, I overlooked the substitution and thought you were just substituting in the modulus of z1 and z2. Seriously honest mistake of mine, nt trying to attack you in any way.Not sure what it is you don't get, but all I've done is replace z1 by 1 and z2 by 1. Simple substitution. If you think year 7 substitution is nonsense, well ... what can I say.
Perhaps you could show me where I said that |z1 - z2| equals |z1|-|z2| .
And 'nonsense' is a non-countable noun, so it can't be preceded by 'a' unless you want to use it as an adjective for a countable noun.
Just wondering isn't it possible that A isn't 90 degrees?Draw a circle with three point A, B, C representing z1, z2, z3.
By inspecting the formula, angle at A is 90. Hence using the semi circle angle property, BC must be a diameter through the origin.
Since z1 can b anywhere, so by inspecting the modulus |z1+z3| formed by adding z1 and z3 , hence > 0 and <2. I don't know how others got root 2.