Mathematics Stack Exchange
Mathematics Stack Exchange

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
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Prove that $\left| \sin z \right| \leq 1$ implies $ \left|\operatorname{Im} z\right| \leq \ln\left(\sqrt{2}+1\right)$

Prove that if $\left| \sin z \right| \leq 1$, then $$ \left|y\right| \leq \ln\left(\sqrt{2}+1\right) .$$ I am not familiar with hyperbolic function before learning complex analysis but in order to deal with questions like this, hyperbolic…
Nighty
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Can we write the real part of this expression?

Assuming $a, b \in \mathbb{R}$ and $b >0$, is there a way we can simplify the real part of this expression? $$ i ab \left(e^{-i \sqrt{-a^4 -i a b}}-e^{i \sqrt{-a^4 +i a b}}\right) $$ In case it matters, it was derived using the $(-\infty, 0)$…
WhatIAm
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How to plot $\{z \in \mathbb{C} : |z-i|>|z+i|\}$

How would I draw the set $\{z \in \mathbb{C} : |z-i|>|z+i|\}$ and $\{z \in \mathbb{C} : |z-i|\not=|z+i|\}$? Im not sure how to solve the second one, and for the first one, I tried squaring both sides and trying to work something out, but I got no…
Derrick
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Geometrical interpretation ($\left| \frac{z-1}{z-i}\right| > 1$, $\arg z < \pi$)

I'd like to ask you about the example below (I have to draw a geometrical interpretation in an argand diagram). $\left| \frac{z-1}{z-i}\right| > 1$, $\arg z < \pi$ What I did so far: $|z-1| > |z-i|$ $|x + yi -1| > |x + yi -i|$ $\sqrt{(x-1)^2 + y^2}…
Pawel Gumiela
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Geometrical interpretation $\frac{1}{6}\pi \le arg\frac{z(1+i)}{-1+i} \le \frac{1}{3}\pi$

I'd like to ask you about the example below (I have to draw a geometrical interpretation in an argand diagram). Am I doing the right thing? $$\frac{\pi}{6}\le \arg\frac{z(1+i)}{-1+i} \le \frac{\pi}{3}.$$ First step: $$\frac{\pi}{6} \le \arg(-iz)…
Pawel Gumiela
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Is $\pi$ a complex number?

I just came across an exam paper the answer of one of the questions about complex numbers says $\pi$ is a complex number. How $\pi$ is represented as a complex number?
user256670
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Equations with complex degree

We all know that an equation of degree $n\in\Bbb N$ has exactly $n$ complex roots, even if we don't know how to find them in closed form. But what do we know about equation with complex degree ? I tried a few ones and I got very different…
AlienRem
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If $w$ be nth root of unity , then $1+2w+3w^2+\dots+nw^{n-1}$ is equal to?

I tried it by letting expression $1+2w+3w^2+\dots+nw^{n-1}= x$ and then multiplying $w$ both sides . I subtracted equation 1 from 2 but it does not seems to help me because i have just started learning this topic and i am weak in properties . Can…
Seth Rollins
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Complex numbers argument

Question: $$z=\frac{a+3i}{2+ai}$$ Show that there is only one value of $a$ for which $\operatorname{arg} z= \frac{\pi}{4}$, and find this value. My attempt: $$\frac{a+3i}{2+ai}\cdot\frac{2-ai}{2-ai}$$ $$=\frac {5a+(6-a^2)i}{4+a^2}$$ $$=\frac…
εν οίδα ότι ουδέν οίδα
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Solving for an exponent?

$1^x = i$ I can't solve it through logs, because $\log 1 = 0$. Does this mean $x$ is undefined?
user26649
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Solving $e^{iz} = 1+i$, where have I gone wrong?

I am trying to solve a question which asks to find all the solutions of $e^{iz}=1+i$. Here is what I have done: $z = x + yi$ $e^{i(x+yi)} = 1+i$ $e^{xi-y}=1+i$ $e^{xi-y}=\sqrt2e^{(\frac\pi4 +2n\pi)i}$ $e^{-y}=\sqrt2$ $y=-\frac12ln2$ $x=\frac\pi4…
abruzzi26
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Definition of $i$

I have a definition of $i$ (here) that has come from observing Newton-Raphson (NR) on $(x^2+c)=0$. The attractors when plotting NR were $\pm \sqrt{c}$, which happen to be the coefficients of $i$ in the solution. Since neither attractor was…
Fred Viole
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If $z$ and $w$ are complex numbers such that $|(z+w)| = |(z-w)|$, prove that $\arg(z)-\arg(w) = \pm (\pi/2)$

If z and w are complex numbers such that $|z+w|$ = $|z-w|$, prove that $\arg(z)-\arg(w)= \pm\pi/2$. Can this be solved algebraically or would a graphic interpretation be better. Both methods would be grateful.
Stratos1
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Difficulties with complex numbers and complex planes

Find the general form of a complex number $z$ such that: i) $z^2$ is a real number (i.e. $\operatorname{Im} z^2 = 0$); ii) $z^2$ is a purely imaginary number (i.e. $\operatorname{Re} z^2 = 0$); iii) $z^2 = \bar z$. I'm not sure how to tackle…
user270494
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How to show $z=1$ is a factor without first finding $\lambda$?

Question: Given that $e^{i \theta}$ is a root to the equation $$z^3 - (\sqrt{3} +1)z^2+\lambda z - 1 =0,$$ where $0< \theta < \pi/2$ and $\lambda \in \mathbb{R}$, show that this equation can be factorized as $$(z^2 - 3z \cos \theta +1) (z-1) =…
troggz
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