Merry Christmas from Math Man (2 Viewers)

math man · Dec 25, 2011

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.

michaeljennings · Dec 25, 2011

This is a terrible present. I would hate to be your child

But Merry Christmas Maths man =)

SpiralFlex · Dec 25, 2011

Merry Christmas Math man from Spiral.

If $e^{i \theta} = \cos \theta+i\sin \theta$

Prove that: $i^i=\frac{1}{\sqrt{e^\pi}}$

Lots of robot love.

math man · Dec 25, 2011

SpiralFlex said:
Merry Christmas Math man from Spiral.

If $e^{i \theta} = \cos \theta+i\sin \theta$

Prove that: $i^i=\frac{1}{\sqrt{e^\pi}}$

Lots of robot love.

Sure brightened up my christmas with an interesting question

math man · Dec 25, 2011

michaeljennings said:
This is a terrible present. I would hate to be your child

But Merry Christmas Maths man =)

yeh i would hate to be my kid too

math man · Dec 25, 2011

Spiral i tweaked your question and made it more interesting

SpiralFlex · Dec 25, 2011

Math man. Am I allowed to use the fact that the points inscribe a square inside the circle?

Spiral is a bit dizzy at the moment.

math man · Dec 25, 2011

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

SpiralFlex · Dec 25, 2011

Excuse my dizziness.

By drawing the diagram (which I will do tomorrow), we can see that by vector rotation,

$\frac{z_{1}}{z_{1}+z_{2}-z_{3}}=\frac{z_{3}(\cos (-45)+i\sin (-45))}{-iz_{3}-z_{3}}$

$=\frac{z_{3}(\cos (-45)+i\sin (-45))}{z_{3}(-1-i)}$

$=\frac{(\cos (-45)+i\sin (-45))}{(-1-i)}$

$=\frac{(\cos (-45)+i\sin (-45))}{(\cos (-135)+i\sin (-135))}$

$=\cos 90 + i\sin 90$

$=i$

Hence the expression is purely imaginary.

SpiralFlex · Dec 25, 2011

Are you sure Part II is not $\sqrt{2}*z_{1}$ ?

math man · Dec 25, 2011

SpiralFlex said:
Are you sure Part II is not $\sqrt{2}*z_{1}$ ?

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..."

math man · Dec 25, 2011

SpiralFlex said:
Merry Christmas Math man from Spiral.

If $e^{i \theta} = \cos \theta+i\sin \theta$

Prove that: $i^i=\frac{1}{\sqrt{e^\pi}}$

Lots of robot love.

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)

SpiralFlex · Dec 25, 2011

math man said:
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..."

Haha. Then you would use the property of Pythagoras and the fact the hypotenuse is $\sqrt{2}$ of the sides. I can't describe it though. Was sleepy. =.=

Technically I have used circle geometry. But I will see if I can prove it in a way that satisfies you.

Since $z_{1}$ bisects the angles of the square, it must be a portion of the diagonal of the square. Through vector transformation, $z\rightarrow cz$ where $c$ is a constant greater than one will enlarge the vector by a factor of $c$ . In this case, to extend the point to lie on the circle, the vector $z_{1}$ must be extended by $\sqrt{1^2+1^2}z_{1}$ (from diagonals of a square), ie. $\sqrt{2}z_{1}$

Also, the question I had was a sub-question, so I removed the "hence". That's why it's restricted to using Euler's formula.

math man · Dec 25, 2011

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 and poof you are done

RealiseNothing · Dec 25, 2011

SpiralFlex said:
Merry Christmas Math man from Spiral.

If $e^{i \theta} = \cos \theta+i\sin \theta$

Prove that: $i^i=\frac{1}{\sqrt{e^\pi}}$

Lots of robot love.

Is this right?

RealiseNothing · Dec 25, 2011

Also Merry Christmas everyone.

IamBread · Dec 25, 2011

RealiseNothing said:
Is this right?

Looks right.

seanieg89 · Dec 25, 2011

Merry xmas BOS!

Umm, I am reasonably sure the geometric situation in mathmans original question is not actually possible, and hence the identity required to prove should never actually hold...maybe I am mistaken.

Here is another question, based on the triangle inequality. It is commonly used as the first step in a proof of the Fundamental Theorem of Algebra:

$$Let $P(z)$ be a nonconstant polynomial with complex coefficients. Show that for every $\epsilon>0$, there exists a real number $R>0$ such that $\left|\frac{1}{P(z)}\right|<\epsilon$ for all $z$ with $|z|>R$.$

Carrotsticks · Dec 25, 2011

seanieg89 said:
Merry xmas BOS!

Umm, I am reasonably sure the geometric situation in mathmans original question is not actually possible, and hence the identity required to prove should never actually hold...maybe I am mistaken.

Here is another question, based on the triangle inequality. It is commonly used as the first step in a proof of the Fundamental Theorem of Algebra:

$$Let $P(z)$ be a nonconstant polynomial with complex coefficients. Show that for every $\epsilon>0$, there exists a real number $R>0$ such that $\left|\frac{1}{P(z)}\right|<\epsilon$ for all $z$ with $|z|>R$.$

Does this proof have to use methods only learnt in MX2? I have a general idea for a proof, but I think it requires the use of the max/min modulus principle (http://en.wikipedia.org/wiki/Maximum_modulus_principle), but that's meant to be something learnt in 2nd Year. I only know of it because I've been watching lectures and reading up on a bit of Complex Analysis for next semester.

EDIT: Woops, this proof does the whole FTA, not just the step you said.

math man · Dec 25, 2011

seanieg89 said:
Merry xmas BOS!

Umm, I am reasonably sure the geometric situation in mathmans original question is not actually possible, and hence the identity required to prove should never actually hold...maybe I am mistaken.

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 easily done.

Merry Christmas from Math Man (2 Viewers)

Member

Attachments

Active Member

Well-Known Member

Member

Member

Member

Well-Known Member

Member

Well-Known Member

Well-Known Member

Member

Member

Well-Known Member

Member

what is that?It is Cowpea

what is that?It is Cowpea

Member

Well-Known Member

Retired

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)