Complex Number questions (1 Viewer)

BenHowe · Mar 2, 2017

There might be a couple, but this is the first...
$\text{For each z in the set of complex numbers, show there are unique real numbers}a\text{and}b\text{such that}z=a+bw$

InteGrand · Mar 2, 2017

BenHowe said:
There might be a couple, but this is the first...
$\text{For each z in the set of complex numbers, show there are unique real numbers}a\text{and}b\text{such that}z=a+bw$

What is w? There won't exist such real a and b if w is real, but if w is non-real, then there will.

BenHowe · Mar 2, 2017

InteGrand said:
What is w? There won't exist such real a and b if w is real, but if w is non-real, then there will.

InteGrand · Mar 2, 2017

BenHowe said:
View attachment 33826

There are various ways to show it. It comes out essentially immediately with a bit of Linear Algebra knowledge.

BenHowe · Mar 2, 2017

Uh I haven't done any linear algebra yet,

BenHowe · Mar 2, 2017

I'm also struggling with using field and order axioms to prove stuff like given

$x+z=y+z\\\text{show}x=y$

InteGrand · Mar 2, 2017

BenHowe said:
I'm also struggling with using field and order axioms to prove stuff like given

$x+z=y+z\\\text{show}x=y$

Add the negative of z to both sides and use associativity of addition.

BenHowe · Mar 2, 2017

How do you like refer to the order/field axioms

InteGrand · Mar 2, 2017

BenHowe said:
How do you like refer to the order/field axioms

Depends which ones. You can say things like "associativity of addition" etc.

BenHowe · Mar 2, 2017

Ahh gotcha tbh I worked it out. Thanks for your help. The thing I don't understand still are partial orders and how u'd use it for a q like this. Also what the curved inequality signs mean.

InteGrand · Mar 2, 2017

BenHowe said:
Ahh gotcha tbh I worked it out. Thanks for your help. The thing I don't understand still are partial orders and how u'd use it for a q like this. Also what the curved inequality signs mean.

View attachment 33827

$\noindent That curved inequality sign is being used to denote a partial order $(\geq )$ on the given set.$

omegadot · Mar 5, 2017

InteGrand said:
There are various ways to show it. It comes out essentially immediately with a bit of Linear Algebra knowledge.

$$\noindent Here is one way to show $a$ and $b$ are unique. Let $u = a + b \omega$. If $a$ and $b$ are not unique, then it is possible to write this complex number as $u = \alpha + \beta \omega$. Here $a,b,\alpha, \beta \in \mathbb{R}$ while $\omega$ is the complex cube root of unity whose imaginary part is positive. Also, let $\omega = w_1 + i w_2$ where $w_1 \in \mathbb{R}$ and $w_2 > 0.$

$\noindent So$

$u = a + b\omega = a + b(w_1 + i w_2) = (a + bw_1) + i b w_2$

$\noindent and$

$u = \alpha + \beta\omega = \alpha + \beta (w_1 + i w_2) = (\alpha + b\beta w_1) + i \beta w_2$

$\noindent Thus$

$(a + bw_1) + i bw_2 = (\alpha + \beta w_1) + i \beta w_2.$

$\noindent On equating real and imaginary parts we have$

$a + b w_1 = \alpha + \beta w_1$ and $b w_2 = \beta w_2.$

$\noindent From $b w_2 = \beta w_2$ we see that $(b - \beta) w_2 = 0$. But since $w_2 > 0$ this can only be true provided $b = \beta$. It therefore follows that $a = \alpha$ showing $a$ and $b$ are unique.$

Complex Number questions (1 Viewer)

BenHowe

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