Questions tagged [complex-numbers]

Questions involving complex numbers, that is numbers of the form $a+bi$ where $i^2=-1$ and $a,b\in\mathbb{R}$.

A complex number is a number in the form $z=a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, or alternatively, $z=r\cdot e^{i\theta}$, with $r$ called the magnitude and $\theta$ called the argument.

The complex conjugate, $\overline z$, is $a-bi$ or $r\cdot e^{-i\theta}$.

Read more about complex numbers and their properties here.

19229 questions
2
votes
3 answers

Let $\alpha,\alpha+1\in\Bbb C$ have modulus $1$. If $n\in\Bbb Z$, and $1+\alpha$ is an $n$-th root of unity, prove that $\alpha$ is also one.

Let $\alpha$ be a complex number such that both $\alpha$ and $\alpha+1$ have modulus $1$. If for a positive integer $n$, $1+\alpha$ is an $n$-th root of unity, then show that $\alpha$ is also an $n$-th root of unity and $n$ is a multiple of $6$. I…
Arthur
  • 2,614
2
votes
0 answers

Sketch the set of complex numbers

$$S=\{z ∈ C : |z + \frac{13i}{(3-2i)}| < 2 ∧ -\frac{2π}{3} ≤ Arg(7z) ≤ -\frac{π}{6}\}$$ I multiplied $\frac{13i}{(3-2i)}$ by $3+2i$, so that's what i got $$S=\{z ∈ C : |z + (-2+3i)| < 2 ∧ -\frac{2π}{3} ≤ Arg(7z) ≤ -\frac{π}{6}\}$$ but I don't…
2
votes
4 answers

Regarding uses of $i$ (square root of $-1$)

Are there any uses of 'square root of $-1$' in practical life ; like in Physics ?
Chaitanya
  • 151
  • 3
2
votes
1 answer

Solution for $|z+3-4i|\leq5$ modulus of a complex number

Here's my solution : \begin{align} &|z+3-4i|\leqq 5\\ &|x+3+(y-4)i|\leqq 5\\ &\sqrt{(x+3)^2+(y-4)^2)}\leqq 5\\ &(x+3)^2+(y-4)^2)\leqq 25 \end{align} And is now drawing a circle with $x=-3, y=4$ and $r=5$. My answer will be all numbers lying on…
Tylzan
  • 21
2
votes
2 answers

Find exact solution for complex exponent polynomial equation

Yesterday I found what I thought to be the typical "99% of readers won't be able to solve it" image, with a simple math question. Despite its looks, it yielded this polynomic equation with the imaginary unit as the exponent. $$b^{\pm i}-b = 3.$$ The…
Mowstyl
  • 21
2
votes
1 answer

Computing product of exponential complex numbers, can you do it in less steps?

I am given: $$\left(2-e^{i \theta}\right)\left(2-e^{-i \theta}\right)$$ Which if you expand you get: $$4 - 2(\cos \theta - i\sin \theta) -2 (\cos \theta + i\sin \theta ) + e^0 = 4 - 4\cos \theta + 1 = 5 - 4\cos \theta.$$ Is there a shortcut I could…
Makogan
  • 3,329
2
votes
1 answer

Why is the **complex locus** of $\mathrm{arg}(z)=\theta$ a half line, and not a line?

I know that the locus of $\mathrm{arg}(z)=\theta$ is a half line with angle $\theta$, but I'm not sure why? I can start the proof: $$ z=x+iy $$ $$ \theta=\mathrm{arg}(z)=\arctan\left(\frac{y}{x}\right) $$ $$ \tan(\theta)=\frac{y}{x}…
david4dev
  • 123
2
votes
1 answer

Does the imaginary part in complex number needs to be the square root of one?

I'm not a mathematician, this question is out of curiosity. When I watch a lot of math explained videos, a lot of the usage for complex number is around operation on them which was visualized in a grid-like with $y$ as the imaginary part and $x$ as…
2
votes
1 answer

Help solve this through complex numbers

I have to do the following problem: If $ax+cy+bz = X, cx+by+az=Y, bx+ay+cz=Z$, then show that $(a^2+b^2+c^2-bc-ca-ab)(x^2+y^2+z^2-yz-zx-xy) = X^2+Y^2+Z^2-YZ-XZ-XY$. It's easy enough through normal Algebra if I pair $X^2-XY = X(X-Y)$ etc., and…
ankush981
  • 2,033
2
votes
0 answers

Strange property of polynomial functions. Is this a typo in the book?

There's one Apostol's Exercise that asks to show that if $f : \mathbb{C} \to \mathbb{C}$ is a polynomial function with real coefficients, then $f(z)=f(i)$ for every complex number $z \in \mathbb{C}$. But I'm certain that this is false, because it…
Gold
  • 26,547
2
votes
3 answers

How to find real part of $x^i$ without the "arg(x)"

I am a novice at complex analysis - more experienced in algebra and calculus (real). To find the real part of $x^i$, I first had to do something: Solve $x^i = e^{iy}$ for y then find $a$ and $b$ using Euler's formula. Take $\ln$ of both sides:…
user1112591
2
votes
2 answers

How to solve $3z - i\overline{z}=7-5i$?

I am trying to solve this equation: $3z - i\overline{z}=7-5i$ and I am stuck. I start with changing $z=a+bi$ and $\overline{z}=a-bi$ $3(a+bi) - i(a-bi)=7-5i$ $3a+3bi - (ia-bi^2)=7-5i$ $3a+3bi - ia+bi^2=7-5i$ $3a+3bi - ia-b=7-5i$ And now I don't…
Ridertvis
  • 381
2
votes
1 answer

Is the solution of this complex numbers system correct?

I have a system of two equations that need to be satisfied simultaneously, where $A,B,x,y\in\mathbb{C}$; they have the form: \begin{eqnarray} Ax=-yB\nonumber\\ B\bar{x}=-\bar{y}A \end{eqnarray} where $\bar{z}$ represents the complex conjugate of a…
2
votes
3 answers

how to solve $|z|^2 + z|z| + 2 = 2i Im(z)$?

this is the process I follow: $z|z| = 2iImz -|z|^2- 2 $ $z \sqrt{Re(z)^2 + Im(z)^2} = 2iy - (Re(z)^2 + Im(z)^2) - 2$ $z\sqrt{x^2 + y^2} =2iy -(x^2+y^2)-2$ $z=(2iy -(x^2+y^2)-2)/\sqrt{x^2 + y^2}$ then i get: $Im(z)=2y/\sqrt{x^2 + y^2}$ and…
2
votes
2 answers

The real and imaginary parts of $\frac{z}{(1-z)^2}$

I have been trying to evaluate the real and imaginary parts of $\frac{z}{(1-z)^2}$ and I was wondering if my solution was correct, as it's quite not nice. I first wrote out the z's as (x+iy)'s, turning the denominator into $1+x^2 -y^2 -2x…
J. Ross
  • 93