10
votes
Accepted
Questions about $\mathbb{C}/n\mathbb{C}$
Note that the expression $\mathbb{Z}/n\mathbb{Z}$ is not some special notation. It is the quotient of the group (resp. ring) $\mathbb{Z}$ by the subgroup (resp. ideal) $n\mathbb{Z}$, which is by ...
- 6,115
4
votes
Accepted
Norm-Square of $1−i+2e^{iπ/4}$
Your $2i(e^{i\pi/4}-e^{-i\pi/4})$ term indeed went poof. Another route to calculate this would be to note that $e^{i\pi/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ so that $1-i+...
- 109
2
votes
Accepted
How to determine argument from Euler’s form of complex numbers?
If $z = a + bi$ then we have
$$\begin{eqnarray} a + bi & = & r e^{i \varphi} \\
& = & r \left(\cos \varphi + i \sin \varphi \right) \\
\implies a & = & r \cos \varphi \\
\mbox{...
- 22.5k
2
votes
How to determine argument from Euler’s form of complex numbers?
For a same specific complex number, changing $r$ will change the number. You can only maintain the number constant by changing its argument (by adding $2 \pi k, k \in \mathbb{Z}$)
Changing $r$ will ...
- 308
2
votes
Accepted
Number of possible polynomials satisfying $[P(x)]^2+[Q(x)]^2=x^{2n}+1$
If $P+iQ$ is divisible by both $r_j$ and $\overline{r_j}$ then
$$
P(z)+iQ(z) = (z-r_j)(z-\overline{r_j}) R(z)
$$
with some (complex) polynomial $R$. Since $P$ and $Q$ have real coefficients is
$$
P(...
- 109k
2
votes
Integral: $\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx$
Take advantage of symmetry, substitute $y=\cos x$, integrate by parts, and fold up the integral to get
$$\begin{align*}
\mathcal I &= \int_0^{2\pi} \arctan \frac{1+2\cos x}{\sqrt3} \, dx \\
&= ...
- 17.7k
2
votes
Proving $\sum_{k=-N}^N e^{ik θ} = \frac{\sin[(N+1/2) θ]}{\sin(θ/2)}$
\begin{align*} \sum_{k=-N}^N e^{ik\theta} & = e^{-iN\theta}\sum_{k=0}^{2N+1} e^{ik\theta}=e^{-iN\theta}\cdot\frac{e^{i(2N+2)\theta}-1}{e^{i\theta}-1}=\frac{e^{i(N+1)\theta}-e^{-iN\theta}}{e^{i\...
- 116
2
votes
Accepted
Are the two expressions $a \neq \pm i\sqrt{2}b$ and $b \neq \pm i \frac{1}{\sqrt{2}}a$ equivalent?
Your reasoning is fine. To see why, expand out the formulas of the form $s \neq \pm t$ to what they actually mean, i.e., $s \neq t \land s \neq -t$, which we can also write as $\lnot (s = t \lor s = -...
- 46.1k
1
vote
Accepted
A complex vector subspace is stable for complex conjugation if and only if it is the complexification of a real vector subspace
You need to take
$$\tilde\phi(v)=\phi\left(\frac{v + c(v)}{2}\right) + i\phi\left(\frac{v - c(v)}{2i}\right)$$
because
$$c\left(\frac{v-c(v)}{2}\right)=\frac{c(v)-c(c(v))}{2}=-\frac{v-c(v)}{2}$$
which ...
- 2,794
1
vote
Accepted
Show $\cos3\theta = 4\cos\theta(\cos\theta -\cos\frac{\pi}{6})(\cos\theta -\cos\frac{5\pi}{6})$
$$\dfrac{ (z^2 + 1)(z^2 - \sqrt{3}z + 1)(z^2 + \sqrt{3}z + 1)}{z^3} = \dfrac{z^2+1}{z}\cdot\dfrac{z^2 - \sqrt{3}z + 1}{z}\cdot \dfrac{z^2 + \sqrt{3}z + 1}{z}$$$$ = \bigg(z+\dfrac1z\bigg)\cdot\left[\...
- 6,496
1
vote
Accepted
Prove that $\lim_{z \rightarrow 1-i} |\bar{z}^{2}-2| = 2\sqrt{2}$
If your intention is to solve it through $\epsilon - \delta$ definition, then observe that:
$$|(z^2-2)-((1-i)^2-2)| =|z^2-(1-i)^2|.$$
Now,
$$|z^2-(1-i)^2| = |z-(1-i)||z+(1-i)|.$$
That is
$$|z^2-(1-i)^...
- 1,538
1
vote
How do I get the square root of a complex number?
A formula that works for $z$ not on the negative real axis:
$$\bbox[5px,border:2px solid red]
{\sqrt{z} = \pm \frac{(z + |z|)}{\sqrt{2(\mathcal{Re}\ {z}+|z|)}}
}$$
Example:
$$\sqrt{3 + 4i}= \pm \frac{...
- 54k
1
vote
Does multiplication by the split-complex number $j$ have a geometric interpretation?
This is more of a comment than an answer, but it's too long. Multiplication by $j$ corresponds to reflecting points across the diagonal (the $y = x$ line) of the plane; $j \cdot (x + j y) = y + j x$.
...
- 1,373
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