## Math Genius: Find \$lim_{nto infty}e^{-i/n}z\$, \$zinmathbb{R}\$

Find $$lim_{nto infty}e^{-i/n}z$$ $$zinmathbb{R}$$

$$e^{-i/n}zto e^0z=z$$ as $$n to infty,$$ since $$exp$$ is continuous.

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## Math Genius: Summation containing \$n\$-th roots of unity

If $$a_{0},a_{1},a_{2},a_{3}….a_{n-1}$$ are the nth roots of unity, then the value of $$sum_{i=0}^{n-1}{left(frac{a_{i}}{3-a_{i}}right)}.$$

I tried expressing $$a_{i}$$ in euler form as $$e^{frac{2kpi i}{n}}$$ but I am stuck after it.

Edit: It is a question from JEE examination.

begin{align} sum_{k=0}^{n-1}frac{e^{frac{2pi ik}{n}}}{3-e^{frac{2pi ik}{n}}} &=sum_{k=0}^{n-1}frac{frac13e^{frac{2pi ik}{n}}}{1-frac13e^{frac{2pi ik}{n}}}tag1\ &=sum_{m=1}^inftysum_{k=0}^{n-1}frac1{3^m}e^{frac{2pi ikm}{n}}tag2\ &=sum_{m=1}^inftyfrac1{3^m}n,[nmid m]tag3\ &=sum_{j=1}^inftyfrac1{3^{jn}}ntag4\[6pt] &=frac{n}{3^n-1}tag5 end{align}
Explanation:
$$(1)$$: cancel $$3$$ in numerator and denominator
$$(2)$$: use the Taylor series $$frac{x}{1-x}=sumlimits_{k=1}^infty x^k$$
$$(3)$$: $$sumlimits_{k=0}^{n-1}e^{frac{2pi ikm}{n}}=n,[nmid m]$$ (Iverson brackets)
$$(4)$$: select the terms where $$m=jn$$
$$(5)$$: sum the geometric series

Explanation of step $$boldsymbol{(3)}$$

Note that the sum of the following finite geometric series
$$sumlimits_{k=0}^{n-1}e^{frac{2pi ikm}{n}}=frac{e^{frac{2pi inm}{n}}-1}{e^{frac{2pi im}{n}}-1}=frac0{e^{frac{2pi im}{n}}-1}tag6$$
shows that if $$nnmid m$$,
$$sumlimits_{k=0}^{n-1}e^{frac{2pi ikm}{n}}=0tag7$$
It is pretty simple to see that if $$nmid m$$,
$$sumlimits_{k=0}^{n-1}e^{frac{2pi ikm}{n}}=sumlimits_{k=0}^{n-1}1=ntag8$$

This is only a hint,

begin{aligned} sum_{i=0}^{n-1}{left(frac{a_{i}}{3-a_{i}}right)}+n&=sum_{i=0}^{n-1}{left(frac{1}{1-frac{a_{i}}{3}}right)}\ &=frac{sum_{i=0}^{n-1}{left(prod_{jneq i}{left(1-frac{a_{j}}{3}right)}right)}}{prod_{i=0}^{n-1}{left(1-frac{a_{i}}{3}right)}} end{aligned}

Then You can use Vieta’s formula.

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## Math Genius: How to show that \$Releft(sqrt[6]{z}right)\$ is \$frac{sqrt{2sqrt{2} +4}}{2}?\$

The complex number $$z = -4sqrt{2} + 4sqrt{2}i$$ is given. Show that $$Releft(sqrt[6]{z}right)$$ is $$cfrac{sqrt{2sqrt{2} +4}}{2}.$$

The previous part of the question asks to convert the $$z$$ into Euler’s form.

I’m able to do so and get:

$$z = 8 , e^{ifrac{3pi}{4}}$$

For $$sqrt[6]{z}$$, I use this form to obtain roots. My reasoning is that the real part of $$sqrt[6]{z}$$ will be the same for all roots, as they’re periodically repeated.

$$sqrt[6]{z} = sqrt[6]{8} ,left(e^{ifrac{3pi}{4}+ 2kpi}right)^{frac{1}{6}}$$
$$iff sqrt[2]{2} , left(e^{ifrac{3pi + 8kpi}{24}}right)$$

It follows that:

$$Re(sqrt[6]{z}) = sqrt[2]{2} , cosleft(cfrac{pi}{8}right) ; text{, for k = 0}$$

How should I go about solving this?

You know thatbegin{align}frac{sqrt2}2&=cosleft(fracpi4right)\&=cosleft(2fracpi8right)\&=cos^2left(fracpi8right)-sin^2left(fracpi8right)\&=2cos^2left(fracpi8right)-1end{align}and therefore$$cosleft(fracpi8right)=sqrt{frac{sqrt2}4+frac12}=frac{sqrt{2+sqrt2}}2.$$

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## Math Genius: How to show that \$Releft(sqrt[6]{z}right)\$ is \$frac{sqrt{2sqrt{2} +4}}{2}?\$

The complex number $$z = -4sqrt{2} + 4sqrt{2}i$$ is given. Show that $$Releft(sqrt[6]{z}right)$$ is $$cfrac{sqrt{2sqrt{2} +4}}{2}.$$

The previous part of the question asks to convert the $$z$$ into Euler’s form.

I’m able to do so and get:

$$z = 8 , e^{ifrac{3pi}{4}}$$

For $$sqrt[6]{z}$$, I use this form to obtain roots. My reasoning is that the real part of $$sqrt[6]{z}$$ will be the same for all roots, as they’re periodically repeated.

$$sqrt[6]{z} = sqrt[6]{8} ,left(e^{ifrac{3pi}{4}+ 2kpi}right)^{frac{1}{6}}$$
$$iff sqrt[2]{2} , left(e^{ifrac{3pi + 8kpi}{24}}right)$$

It follows that:

$$Re(sqrt[6]{z}) = sqrt[2]{2} , cosleft(cfrac{pi}{8}right) ; text{, for k = 0}$$

How should I go about solving this?

You know thatbegin{align}frac{sqrt2}2&=cosleft(fracpi4right)\&=cosleft(2fracpi8right)\&=cos^2left(fracpi8right)-sin^2left(fracpi8right)\&=2cos^2left(fracpi8right)-1end{align}and therefore$$cosleft(fracpi8right)=sqrt{frac{sqrt2}4+frac12}=frac{sqrt{2+sqrt2}}2.$$

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## Math Genius: Prove a unique real matrix exists to denote complex numbers in the set of Cauchy–Riemann matrices

One model for the complex numbers $$mathbb{C}$$ uses the set of Cauchy–Riemann matrices

$$CR:=begin{pmatrix}a&b\c&dend{pmatrix}$$ such that $$a,b,c,d in mathbb{R}, a=d, b+c=0$$ with matrix addition and matrix multiplication corresponding to $$+$$ and $$×$$ in $$mathbb{C}$$. We say that a matrix$$begin{pmatrix}a&b\c&dend{pmatrix} in$$ CR is real if $$b=c=0$$ and imaginary if $$a=d=0$$, we write $$i:=begin{pmatrix}0&-1\1&0end{pmatrix}$$

$$(a)$$ Show that for every imaginary $$z ∈$$ CR there is a unique real matrix $$u ∈$$ CR such that $$z = u · i.$$

$$(b)$$ Show that for every $$z ∈$$ CR there are unique real matrices $$u, v ∈$$ CR such that $$z = u + v · i$$.

I believe for both parts of the question, I am missing a key aspect of proving uniqueness. Any help would be appreciated

Since $$i^2=-I_2$$, For (a) the required $$u$$ is $$-zi=-left(begin{array}{cc} 0 & b_{z}\ -b_{z} & 0 end{array}right)left(begin{array}{cc} 0 & -1\ 1 & 0 end{array}right)=left(begin{array}{cc} -b_{z} & 0\ 0 & -b_{z} end{array}right).$$For (b) note that$$u+vi=left(begin{array}{cc} a_{u} & 0\ 0 & a_{u} end{array}right)+left(begin{array}{cc} a_{v} & 0\ 0 & a_{v} end{array}right)left(begin{array}{cc} 0 & -1\ 1 & 0 end{array}right)=left(begin{array}{cc} a_{u} & -a_{v}\ a_{v} & a_{u} end{array}right),$$which agrees with $$left(begin{array}{cc} a_{z} & b_{z}\ -b_{z} & a_{z} end{array}right)$$ iff $$a_{u}=a_{z},,a_{v}=-b_{z}$$.

Some intuition on CR matrices may be gained via inspection of the following simple matrix equations (1)-(3), easily verified:

$$begin{bmatrix} a & 0 \ 0 & a end{bmatrix} begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = begin{bmatrix} 0 & -a \ a & 0 end{bmatrix}; tag 1$$

$$begin{bmatrix} 0 & -a \ a & 0 end{bmatrix}begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} = begin{bmatrix} -a & 0 \ 0 & -a end{bmatrix}; tag 2$$

$$begin{bmatrix} 0 & -a \ a & 0 end{bmatrix}begin{bmatrix} 0 & 1 \ -1 & 0 end{bmatrix} = begin{bmatrix} a & 0 \ 0 & a end{bmatrix}; tag 3$$

we have

$$i = begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag 4$$

$$i^2 = -I; tag 5$$

$$(-i)i = -i^2 = I; tag 6$$

$$i^{-1} = -i. tag 7$$

Note that (1) shows that for every imaginary $$z in CR$$ there is a real $$u in CR$$ such that

$$ui = z; tag 8$$

if

$$yi = z, tag 9$$

then in light of (5),

$$-y = -yI = yi^2 = zi, tag{10}$$

or

$$y = -zi, tag{11}$$

which shows that there is precisely one $$u$$ satisfying (8); thus is item (a) resolved.

We next observe that the only $$w in CR$$ which is both real and imaginary is $$0$$, for if $$w$$ is real then

$$w = begin{bmatrix} a & 0 \ 0 & a end{bmatrix} tag{12}$$

for some $$a in Bbb R$$, and since $$w$$ is also imaginary, we have

$$w = begin{bmatrix} 0 & b \ -b & 0 end{bmatrix} tag{13}$$

for some $$b in Bbb R$$; equating these two forms of $$w$$ yields

$$begin{bmatrix} a & 0 \ 0 & a end{bmatrix} = begin{bmatrix} 0 & b \ -b & 0 end{bmatrix}; tag{14}$$

comparing entries of these two matrices indicates that

$$a = b = 0, tag{15}$$

whence

$$w = 0, tag{16}$$

as asserted.

Now any $$z in CR$$ may be written

$$z = begin{bmatrix} a & -b \ b & a end{bmatrix} tag{17}$$

for some $$a, b in Bbb R$$; we have

$$z = begin{bmatrix} a & 0 \ 0 & a end{bmatrix} + begin{bmatrix} 0 & -b \ b & 0 end{bmatrix} = begin{bmatrix} a & 0 \ 0 & a end{bmatrix} + begin{bmatrix} b & 0 \ 0 & b end{bmatrix}begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix}; tag{18}$$

we denote the real matrices occurring in this equation by $$u$$ and $$v$$:

$$u = begin{bmatrix} a & 0 \ 0 & a end{bmatrix}, tag{19}$$

$$v = begin{bmatrix} b & 0 \ 0 & b end{bmatrix}, tag{20}$$

then using (4) we have

$$z = u + vi; tag{21}$$

if there were two other real matrices $$u’$$ and $$v’$$ such that

$$z = u’ + v’i, tag{22}$$

then

$$u’ + v’i = u + vi, tag{23}$$

or

$$u’ – u = vi – v’i = (v – v’)i; tag{24}$$

since $$u’ – u$$ and $$v – v’$$ are both real, $$(v – v’)i$$ is imaginary in light of (1), and thus by what we have just seen in (12)-(18) it follows that

$$u’ – u = (v – v’)i = 0, tag{25}$$

whence

$$u’ = u, tag{26}$$

and again by virtue of (5),

$$v – v’ = (v – v’)I = -(v – v’)i^2 = (-(v – v’)i)i = 0, tag{27}$$

and thus

$$v = v’ tag{28}$$

as well. Thus the uniqueness of $$u$$, $$v$$ as in (21) is established, and we have dispensed with item (b).

## Math Genius: How to show that \$Releft(sqrt[6]{z}right)\$ is \$frac{sqrt{2sqrt{2} +4}}{2}?\$

The complex number $$z = -4sqrt{2} + 4sqrt{2}i$$ is given. Show that $$Releft(sqrt[6]{z}right)$$ is $$cfrac{sqrt{2sqrt{2} +4}}{2}.$$

The previous part of the question asks to convert the $$z$$ into Euler’s form.

I’m able to do so and get:

$$z = 8 , e^{ifrac{3pi}{4}}$$

For $$sqrt[6]{z}$$, I use this form to obtain roots. My reasoning is that the real part of $$sqrt[6]{z}$$ will be the same for all roots, as they’re periodically repeated.

$$sqrt[6]{z} = sqrt[6]{8} ,left(e^{ifrac{3pi}{4}+ 2kpi}right)^{frac{1}{6}}$$
$$iff sqrt[2]{2} , left(e^{ifrac{3pi + 8kpi}{24}}right)$$

It follows that:

$$Re(sqrt[6]{z}) = sqrt[2]{2} , cosleft(cfrac{pi}{8}right) ; text{, for k = 0}$$

How should I go about solving this?

You know thatbegin{align}frac{sqrt2}2&=cosleft(fracpi4right)\&=cosleft(2fracpi8right)\&=cos^2left(fracpi8right)-sin^2left(fracpi8right)\&=2cos^2left(fracpi8right)-1end{align}and therefore$$cosleft(fracpi8right)=sqrt{frac{sqrt2}4+frac12}=frac{sqrt{2+sqrt2}}2.$$

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## Math Genius: How to show that \$Releft(sqrt[6]{z}right)\$ is \$frac{sqrt{2sqrt{2} +4}}{2}?\$

The complex number $$z = -4sqrt{2} + 4sqrt{2}i$$ is given. Show that $$Releft(sqrt[6]{z}right)$$ is $$cfrac{sqrt{2sqrt{2} +4}}{2}.$$

The previous part of the question asks to convert the $$z$$ into Euler’s form.

I’m able to do so and get:

$$z = 8 , e^{ifrac{3pi}{4}}$$

For $$sqrt[6]{z}$$, I use this form to obtain roots. My reasoning is that the real part of $$sqrt[6]{z}$$ will be the same for all roots, as they’re periodically repeated.

$$sqrt[6]{z} = sqrt[6]{8} ,left(e^{ifrac{3pi}{4}+ 2kpi}right)^{frac{1}{6}}$$
$$iff sqrt[2]{2} , left(e^{ifrac{3pi + 8kpi}{24}}right)$$

It follows that:

$$Re(sqrt[6]{z}) = sqrt[2]{2} , cosleft(cfrac{pi}{8}right) ; text{, for k = 0}$$

How should I go about solving this?

You know thatbegin{align}frac{sqrt2}2&=cosleft(fracpi4right)\&=cosleft(2fracpi8right)\&=cos^2left(fracpi8right)-sin^2left(fracpi8right)\&=2cos^2left(fracpi8right)-1end{align}and therefore$$cosleft(fracpi8right)=sqrt{frac{sqrt2}4+frac12}=frac{sqrt{2+sqrt2}}2.$$

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## Math Genius: Change in argument of sum of complex numbers along a closed path

Let $$a(t)$$ and $$b(t)$$ be complex variables that travel along closed paths that end up at their original points, i.e. $$a(0)=a(T)=a_0$$ and $$b(0)=b(T)=b_0$$, with $$Delta Arg(a) = Delta Arg(b) = 0$$ over the paths. Is it possible for their sum $$c(t) = a(t) + b(t)$$ to have $$Delta Arg(a) = 2pi k$$ where $$k$$ is a non-zero integer? Any example? Also, let’s assume $$a(t) neq 0, b(t) neq 0$$, and $$c(t) neq 0$$ for any $$0 leq t leq T$$.

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## Math Genius: Are there any numbers more fundamental than Complex numbers?

Are there any numbers other than Complex Numbers which contain something more that Real and Imaginary numbers?

Yes. There is in fact a well-structured set of numbers of dimension \$2^n\$ for all \$n\$, but after a certain point it simply becomes pointless and unwieldy. The next set of numbers is called the quaternions of the form \$a+bi+cj+dk\$, where \$a,b,c,dinmathbb{R}\$ and \$i,j,k\$ are all square roots of \$-1\$. In the quaternions, \$ij=k, jk=i, ki=j\$ but \$ji=-k, kj=-i, ik=-j\$ (that’s right, they’re non-commutative). The complex numbers \$mathbb{C}\$, though, are considered more fundamental than the quaternions \$mathbb{H}\$ or the sets that follow (the octonions, sedenions, etc.), since the complex numbers are an abelian field with algebraic closure (algebraic closure of a field \$F\$ means that all polynomials \$sum a_ix^i\$ with \$a_iin F\$ have all of their roots in \$F\$. This cannot be said of \$mathbb{R}\$, since \$x^2+1\$ has real coefficients but imaginary roots.), giving it the greatest structure of all of these number sets. Funny things start to happen after the quaternions, as the octonions are no longer associative and the sedenions can have zero divisors (which means two non-zero elements \$a\$ and \$b\$ can have the property \$ab=0\$).

The Quaternions \$mathbb H\$ maybe? They are non-commutative, though.

The Cayley Dickson construction might be a useful related remark.

Going off in a slightly different direction than sets containing the complex numbers are the hyperreals \${}^astmathbb{R}\$, which is used in non-standard analysis. It can be thought of all adding infinitesimals (numbers that are unequal to zero but smaller in magnitude then any non-zero real number) and infinitely large numbers (larger in magnitude than any number). The are the rigorous formulation of the intuitions using infinitesimals and indivisibles before the \$epsilon-delta\$-definitions were used to bypass the limited rigor employed at the time. What’s nice about this is that it remains a field, and is in a sense equivalent to the reals through the “Transfer principle”, which relates first-order sentences in \${}^astmathbb{R}\$ to those in \$mathbb{R}\$.

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## Math Genius: Will the image of the points inside a circle be inside the image of the circle after this transformation in the complex plane?

In the image below (part (b)), Since $$z < |3|$$ before the transformation, does that simply imply that the region to be shaded after the transformation is definitely the inside of the circle and not its outside? It seems so to me but I don’t want to jump to the conclusion.

Also is the convention to use dotted lines for the locus of the circle, since $$z < |3|$$ as opposed to $$z leq |3|$$ ?

If \$|z – (-i)|\$ is very small, then \$|w|\$ will be very large. In fact, within a certain neighborhood of \$-i\$, the closer \$z\$ is to \$-i\$, the larger \$|w|\$ is. This implies that wherever the circle \$C\$ is, there will be points in a neighborhood of \$-i\$ that are mapped to points outside the circle \$C\$. But \$|z| < 3\$ for at least some of those points.

So it seems not to be a good assumption that the image of the points inside a circle will be inside the image of the circle. In this case, in fact, they are not.

I’m not sure if you still want information on this question but I was recently stuck on a similar problem. I presume you were fine with part a), and if my working is correct I get the following equation for the circle in the \$w\$-plane:

\$\$
(u-frac{9}{8})^2 + v^2 = (frac{3}{8})^2
\$\$

This describes a circle with centre (\$frac{9}{8}\$, 0) and radius \$frac{3}{8}\$.

To get to this answer part of my working involved rearranging the given equation to make \$z\$ the subject and then taking the modulus of either side:

\$\$
|z| = frac{|w|}{|1-w|}
\$\$

From this step it is simple to replace the \$|z|\$ with \$3\$ as we are told this is what the modulus of \$z\$ is. But in part b) we are told that \$|z|<3\$. So the equation becomes:

\$\$
3 > frac{|w|}{|1-w|}
\$\$

Following my working for a), I arrived at the conclusion that:

\$\$
(u-frac{9}{8})^2 + v^2 > (frac{3}{8})^2
\$\$

Hence it would seem that the new region to shade is outside the cirlce \$C\$ because the coordinates of any point in this region must lie a distance greater than \$frac{3}{8}\$ from the circle centre.

Hope this is useful!

(With regards to dotted lines, I agree with your thinking!).

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