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yeh im doing adv complex analysis next yr..can't wait lol

yeh this is the method i posted before...but i did forget to say pv...however these young minded 4u students need not worry about logarithms of complex numbers and the periodicity of the exponential function

Well you may be right for part ii)... root2 z1 may not lie on the original circle, however it does lie on another cirlce....but this is only a simple 4u question so it does not matter if it does or does not lie on the original circle....you just need to prove it lies on a circle, which is...

well i can use that method using eulers formula..i just add an extra step or 2...hmmm im tossing and turning whether to give you part ii) right or not..what you are meant to do is prove that z2 + z3 lies on the circle, which is very easy, then you sub in z2 and z3 for expressions involving z1...

also i noticed the way you proved this question...which i think is mediocre and not in the true fashion of turning LHS/RHS into the other side...but my way uses the logarithm of complex numbers...making the question easy.. (also notice how euler's formula is not necessary here)

i didnt see i still had the 2 there...yes root2 is right...and part i is technically not right as z3 = (cis45)z1 not cis-45...and you were meant to use circle geo to do it really easy as if you read part i) closely i use the words "if...are concylic points prove..."

well i dont wanna give it away..but it does have something to do with a square

Spiral i tweaked your question and made it more interesting

yeh i would hate to be my kid too :P

Sure brightened up my christmas with an interesting question

As it is christmas today i thought why not celebrate it by doing some maths.. So my present to you is this question i wrote. I hope you enjoy it.

it is sort of table of values...but you dont actually write it down...its all just a mental process

A fast method is to appraoch it like this.... y=- sinx exists between -1 and 1 and now all exponentials are > 0 for all x and as the max value is 1 the max value on the composite graph will be 2^1=2....now when -sinx = 0 that will become 1 on the composite and when -sinx = -1 that will become...

yeh i did...i guess doing this post at 3 am was not a good idea

this question is quickly solved geometrically by using vector addition.....as |z| = |w| we will form a rhombus and hence the diagonals z-w and z+w intersect at 90 degrees....therefore arg[(z+w)/(z-w)] is 90 and hence (z+w)/(z-w) is purely imaginary....to prove this algebraicly....times top and...

since OAB is equilateral z2 = cis60 z1...now this is a proof question so taking the LHS= z1^2 + z2^2 this becomes z1^2 + (cis60z1)^2...Simplyfying further...LHS= z1^2 + cis120z1^2=z1^2(1+cis120)=z1^2cis60 as if you expand 1 + cis120 you will get cis60... therefore LHS = z1(z1cis60) = z1 z2 = RHS

here is the geometric method

Referring to diagram above let M be midpoint of two diagonals.....Now spiral already proved part b the most efficient way which everyone should do...so for part c angle CAD = x/2 as diagonals of a rhombus bisect the vertice angles. angle AMC is 90 as diagonals of rhombus bisect at 90. Now as...

you are forgetting that |z1-z2| also forms one of the sides of a triangle when you use the parallelogram of addition for vectors, hence the triangle inequality can be used for |z1-z2| as well as |z1+z2|

there are two forms of the triangle inequality which you frequently use to find min and max mod of |z1 + z2|... the first form which everyone should know is : |z1 + z2| <= |z1| + |z2| which geometrically means that the third side of a triangle is less than the sum of the other two sides and...