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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Help with complex number phasor notation

I am having trouble understanding how $10jy$ is converted to $10 e^{j\pi/2}$. Here $x$ and $y$ are unit vectors: (original image) $$\large=\operatorname{Re}\left[(10\hat{x}-10j\hat{y})e^{-j10\pi…
GorillaApe
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Prove that if $z+\frac1z$ is real, then either $|z|=1$ or $z$ is real.

Prove that if $z+\frac1z$ is real, then either $|z|=1$ or $z$ is real. Original image I am not sure whether my proof is sufficient. So far, I have shown that $$z+\frac1z = \frac{z^2+1}{z}=\frac{|z|+1}{z}$$ However, I don't think the proof…
CCC
  • 1,031
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Solving equation involving complex numbers

(a) Find real numbers $a$ and $b$ such that $(a+bi)^2 = -3-4i.$ (b) Hence solve the equation: $z^2+i\sqrt{3}z+i = 0$. Original Image In the above question, I have solved part (a), with $a=\pm1$ and with $b=\mp2$, but I am not sure how to use this…
CCC
  • 1,031
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Find $i\log(x-i)+i^2\pi+i^3\log(x+i)+i^4(2\arctan x)$, if $x>0$

Find $i\log(x-i)+i^2\pi+i^3\log(x+i)+i^4(2\arctan x)$, if $x>0$ The equation can be written as $$y=i\log(x-i)-\pi-i\log(x+i)+2\arctan x$$ $$y=i\log\frac{x-i}{x+i}-\pi+2\arctan x$$ Let $x+i=re^{i\theta}$…
Aditya Dev
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If $z$ and $\omega$ are two complex no. and $\theta = \arg\left(\frac{\omega -z}{z}\right)\;,$ Then Max. of $\tan^2 \theta$

If $\omega$ and $z$ are two complex number such that $|\omega| = 1$ and $|z|=10$ and Let $\displaystyle \theta = \arg\left(\frac{\omega -z}{z}\right)$ Then Maximum possible value of $\tan^2 \theta$ $\bf{My\; Try::}$ Let $\omega = =e^{i\alpha} =…
juantheron
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Is there any difference between the absolute values operators $|z|$ and $\|z\|$?

Is there any difference between the absolute values operators $|z|$ and $\|z\|$ where $z=a+ib$?
user29646
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2 answers

Complex Numbers (Locus)

Find the Cartesian equation of $z$ described by $$\arg\left(\frac{z-2}{z+5}\right)=\frac{\pi}{4}$$ So what I have done is let $z = x+iy$ $$\frac{z-2}{z+5} $$ $$ \frac{x+iy-2}{x+iy+5} $$ $$ \frac{x-2+iy}{x+5+iy} \frac{x+5-iy}{x+5-iy} $$(rationalizing…
bigfocalchord
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How to find the minimum value sum of three complex numbers given constrains?

Question) If $|z_1 -1|<1$, $|z_2 -2|<2$, $|z_3 -3|<3$ then $|z_1+z_2+z_3|$ is: (a) is less than 6 (b) is more than 3 (c) is less than 12 (d) lies between 6 and 12 My Attempt: (1) Using polygon inequality…
Sujith Sizon
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Simplify the complex expression to complex numbers (2i)^i

I am getting stuck with this one.... This is how far I got $$2i = 2 \exp\left(i\frac{\pi}{2} + 2k\pi\right)$$ $$(2i)^I = 2^{\frac{i}{2}}$$ I'm having trouble moving past that point
Treshon
  • 23
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Find z3 in the form of x+jy

I would need to find 3z in the form of x+jy where: $$\frac{1}{3z}=\frac{1}{(3-j4)}+\frac{1}{(3-j4)(5+j2)}$$ What I did was to expand the $$\frac{1}{(3-j4)(5+j2)}$$ which gives me $$\frac{1}{(23-14j)}$$ From here I am not very sure how to continue as…
Ong
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solution to a complex equation

I'm asked to solve this equation : $$z^4+7(1+i)z^2+25i=0$$ by taking $z^2 = u $ : $$u^2+7(1+i)u+25i=0$$ this turns out to be a quadratic equation. solving it gives: $$u_1=-6-8i$$ $$u_2=-8-6i$$ so how are we supposed to solve $z^2=-6-8i$ and…
Soroush
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Complex Numbers: $1-z+z^2-z^3=0$

How to solve this equation? ($z$ is a complex number) $1-z+z^2-z^3=0$ I tried using $z=a+ib$ and reached an answer but I'm not sure if it's a correct one. Thanks!
user293412
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Cauchy–Schwarz inequality for complex numbers

How can I prove the Cauchy– Schwarz inequality for two complex numbers? $$z_1=x_1+iy_1$$ $$z_2=x_2+iy_2$$ I can prove the triangle inequality for two complex numbers: $$|z_1+z_2|\le |z_1|+|z_2|.$$ But I cannot prove the Cauchy–Schwarz…
user29646
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2 answers

$z_1,...z_n$ are the $n$ solutions of $z^n =a$ and $a$ is real number, show that $z_1+...+z_n$ is a real number

I was actually trying this simple question, might be just me being really rusty not doing maths for a very long time. I tried adding all the $\theta$ up on all zs but it doesn't seem to work. Anyway as the title states, $z_1,...z_n$ are the $n$…
Yau Kin Hoe
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Is it necessary to proof that $|a-b|$ is equals to $|b-a|$ in $\mathbb{C}$ if we know this is true in $\mathbb{R}$?

I just want to know if it's necessary, I'm not asking for the proof. Can we skip the proof of it if we know that's true in $\mathbb{R}$?
GniruT
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