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.

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Can i make this solution less messy (complex numbers in trigonometric form)

I have the complex roots $x_1=1+3i, x_2=1-3i$ and i want to express them in trig form Since the trig form is $p(cosθ + sinθ)$ and $p=\sqrt{a^2 + b^2}, tanθ=\frac{b}{a}$ we have $p=\sqrt{1^2 + 3^2}=\sqrt{10}$ and $tanθ = \frac{3}{1}=3$ Issue is, that…
Than1
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Let $k$ be a positive integer, $S$ is a set of $k$ nonzero complex numbers, sastifying $a,b\in S\Rightarrow ab\in S$

Let $k$ be a positive integer, $S$ is a set of $k$ nonzero complex numbers, sastifying $a,b\in S\Rightarrow ab\in S$. Show $\forall\ a\in S, a^k=1$, and find the sum of elements of $S$. Clearly, $a,a^2,\dots,a^{k+1}\in S$, and hence for some $i
xldd
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Solving $\cos z=i\sin z$, $z$ complex number

Solving $\cos z=i\sin z$, $z$ complex number. Well, for me is not solution because after replace $$\frac{e^{iz}+e^{-iz}}{2}=i\frac{e^{iz}-e^{-iz}}{2i}$$ equivalently $$2e^{-iz}=0$$ but here i dont find solution. Is fine my attempt? Thank you
weymar andres
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Division algorithm for complex polynomials

The Question: This question is not actually about the complete proof for it, just one step. However, I still feel this is an appropriate title. The question is simply to show that given a polynomial $p(z)$ of degree at least 1, there exists a…
Peatherfed
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If $\overline{z}=-z$, then $z\cdot z<0$

I am currently reading this paper on the axioms of complex numbers. The set of complex numbers is characterized (and defined up to a natural isomorphism) by the following axioms: $\mathbf{C}$ is a commutative field. $\mathbf{C}\ni…
Filippo
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Writing complex numbers in polar form

$z=4\mathrm{e}^{i1.7}$ Write the following numbers in the polar form $r\mathrm{e}^{i\phi}$: a) $\bar{z}$ b) $\displaystyle \frac{1}{z}$ I need $r$ and $\phi$. I already have $r$ for the both of these (a is $4$, b is $.25$) but I'm not sure what…
calicojack
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Prove that the complex number system does not satisfy three properties

From my textbook, The definition of the order relation denoted by > in the real number system is based upon the existence of a subset $P$ (the positive reals) having the following properties. i) For any number $a\ne0$, either $a$ or $-a$ but not…
matt
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Showing that for complex numbers $z_1,z_2,z_3$, $|z_1z_2+z_2z_3+z_3z_1|=|z_1+z_2+z_3|$ given that $|z_i|=1$.

Let $z_1,z_2$ and $z_3$ be complex numbers such that $|z_i|=1$ for $i=1,2,3$. Show that $|z_1z_2+z_2z_3+z_3z_1|=|z_1+z_2+z_3|$ This is an exercise from Jonathan S. Golan's book on Linear Algebra. To check that it is true for particular values of…
tmaj
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A different approach to the question: Determine max $|z^3 − z + 2|$. $z∈\Bbb C, |z|=1 $

Determine max $|z^3 − z + 2|$. $z∈\Bbb C, |z|=1 $ Now, quite clearly, substituting $z = x + iy$ makes the question easy. The solution is not very difficult after that. However, is there some other (shorter or longer) method to solve it and is a…
mathx
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What is the solution of equation $z^2 - 2 \bar{z} = 3$?

my question is how to solve the equation $z^2 - 2 \bar{z} = 3$, where $z \in \mathbb{C}$? I know that polar form of any complex number is $z = r \cos(\varphi) + i \cdot r \sin (\varphi)$. And similarly $\bar{z} = r \cos(\varphi) - i \cdot r…
andrej benedičič
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Find $x$ and $y$ such that $(x+3i)(x+iy)=1-i$

I tried expanding the brackets, equating Real and Imaginary parts and finally substituting $x$ in terms of $y$. I ended up with the equation: $$3y^3 + 19y^2 + 33y +8=0$$ Using a calculator, I got a root at $y\simeq-0.288$ which is approx.…
puzzlD
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Complex solutions of $z^4 = |z|^2 +2$

I'm struggling with the equation in the subject, I managed to find the real solutions ($\pm\sqrt{2}$) by setting $u = z^2$ and I know from Wolfram Alpha that the other two complex solutions should be $\pm i\sqrt{2}$ and I obviouvsly understand why…
complexguest
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The complex equation: $z^4 + 1 = \dfrac{1 - iz^3}{1 - z^4}$ How to solve it in $\mathbb{C}$ numbers?

I have here an complex equation in $\mathbb{C}$ numbers: $$z^4 + 1 = \dfrac{1 - iz^3}{1 - z^4}.$$ My question is how to solve this complex equation? What is the best way to solve it? My attempt: Multiply both sides by $1-z^4$ to get…
Andrej
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If $A(z_1) , B(z_2) , C(z_3)$ are complex numbers satisfying $|\mathrm{z}-\sqrt{3} \mathrm{i}|=1$ and .... Find $\left|z_{1}-z_{2}\right|$

If $A(z_1) , B(z_2) , C(z_3)$ are complex numbers satisfying $|\mathrm{z}-\sqrt{3} \mathrm{i}|=1$ and $3 \mathrm{z}_{1}+\sqrt{3} \mathrm{i}=2 \mathrm{z}_{2}+2 \mathrm{z}_{3}$. Find $\left|z_{1}-z_{2}\right|$ My Approach: I can more or less just…
marks_404
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How to reconcile the identity $\left( e^{i \theta} \right)^{1/2} = e^{i \theta/2}$ with the fact that $a^2 = b$ is solved by $a = \pm\sqrt{b}$

So I worked along the lines of the following: $$ \left( \cos \left( \theta \right) + i \sin \left( \theta \right) \right)^{\alpha} = \left( e^{i \theta} \right)^{\alpha} = e^{i (\theta \alpha)} = \cos \left( \theta \alpha \right) + i \sin \left(…
DanZimm
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