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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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For complex $z$, find the value of $z$; $z^2 -2z+( 3-4i) =( 6+3i)$

\begin{array}{l} \boldsymbol{z^{2} -2z+( 3-4i) =( 6+3i)}\\\\ z^{2} -2z+( 3-4i) =( 6+3i)\\ x^{2} -y^{2} +2xyi-2x-2yi-3-7i=0\\ \left( x^{2} -y^{2} -2x-3\right) +i( 2xy-2y-7) =0\\ x^{2} -y^{2} -2x-3=0\rightarrow ( 1)\\ 2xy-2y-7=0\rightarrow ( 2)\\ ( 2)…
RGen
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How do I express a complex number in a complex base?

I came across an old mathematical paper on the web, published in the 1980s (I can't seem to find it again). The paper was about complex number arithmetic, and it talked about expressing complex numbers in complex bases. I would like to express a…
John Doe
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Confusion Regarding General Form Of Triangle Inequality For Complex Numbers

My textbook has the following to say about the Triangle Inequality: Now my question is regarding the general form. It says that the equality sign holds iff the ratio of any 2 non-zero terms is positive. But what does that mean? Is it talking about…
Anili
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If $(1+x+x^2)^n = (p_0+p_1x+p_0x^2+p_3x^3+p_4x^4+\ldots++p_nx^n)$ then prove that $p_1+p4+p7 + \ldots = 3^{n-1}$

I have the following questins: If $(1+x+x^2)^n = (p_0+p_1x+p_0x^2+p_3x^3+p_4x^4+\ldots++p_nx^n)$ then prove that $p_1+p4+p7 + \ldots = 3^{n-1}$ I have taken x = 1, $\omega$, $\omega^2$ and then sum all the equations. By this, I am able to prove…
Christopher Marlowe
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Find all number $z$ such that $\cos(z) = i\sin(z)$.

Find all number $z$ such that $\cos(z) = i\sin(z)$. Options : $$(a) \ in, \ n\in \mathbb{Z} \\ (b) \ inπ +1, \ n\in \mathbb{Z} \\ (c) \ \frac{inπ}{2}, \ n\in \mathbb{Z} \\ (d) \ \text{no solutions exist}.$$ My try : Let $z=x+iy$. Then, $\cos(z)…
Itachi
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Is there any possibility to get $\Gamma \subset \Lambda$?

Let $f,g$ be $n$-degree and $m$-degree polynomials over $\mathbb C$ respectively, with $n>m$. Set $\Lambda=\{x \in \mathbb C~:~ f(x)=0 \}$ and $\Gamma=\{x \in \mathbb C~:~ g(x)=0 \}$. Assume $h \circ f=g \circ h$, where $h:~\Lambda \to \Gamma$ is…
MAS
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Prove that there exists a complex number $z$ such that $|z|>1$ and $z^{135} + (2+3i)z -100=0$.

Prove that there exists a complex number $z$ such that $|z|>1$ and $z^{135} + (2+3i)z -100=0$. I understand that we can express $z$ in terms of $r(\cos \theta + i\sin \theta) = re^{i\theta}$ but I honestly don't see how this helps in solving this…
IAAW
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How to show that $f(z) = f(i)$

Exercise problem. I do not need a full solution because I am trying to solve myself. Just a hint would be great. Let $f$ be a polynomial with real coefficients. How to show that $f(z) = f(i)$ for every complex $z$?
fade2black
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By taking the complex conjugate of the equation, show that the solutions to $z^2 + az + b$ are either real or come in complex conjugate pairs.

I don't understand how you would take the conjugate of a quadratic equation and how it would be useful to solve this question. I would normally show it by saying if $b$ is real, then it is equal to $\alpha$ times $\beta$, say $a$ is equal to…
Xemor
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Are this passages wrong in order to solve a complex equation?

Let's consider the equation in the complex variable $w$ $$\alpha+\frac{w}{4\pi}+\frac{e^{-2\beta w}}{8\pi\beta}=0$$ with $\alpha$ and $\beta$ real parameters, $\beta>0$. By writing $w=a+ib$, it can be put in the form $$\frac{e^{-2\beta…
SpuriousMatemagician
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solve the equation $\left(z^{2}+1\right)^{n}-(z-1)^{2 n}=0$

Im trying to solve the eq $$(E)\quad \left(z^{2}+1\right)^{n}-(z-1)^{2 n}=0$$ My attemp : By Newton i get $$\left(z^{2}+1\right)^{n}-(z-1)^{2n}=0 \Leftrightarrow \sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) z^{2…
Donnie Darko
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solve the equation $1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$

im trying to solve the equation $$(E)\quad 1+2z+2z^2+\ldots +2z^{n-1}+z^n=0$$ attempt : because $1$ isnt a solution we have $$\begin{aligned} 1+2 z+2 z^{2}+\cdots+2 z^{n-1}+z^{n} &=2\left(z^{0}+z+z^{2}+\cdots+z^{n-1}\right)-1+z^{n} \\ &=2…
Donnie Darko
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Finding all four values of $\sqrt{i}+\sqrt{-i}$

$$\begin{align} \sqrt i+\sqrt{-i} & =e^{\frac{i\pi}{2}}+e^{\frac{3i\pi}{2}}\\ &=e^{\frac{i\pi}{2}+\frac{3i\pi}{2}}\\ &=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\\ &=i\sqrt{2} \end{align} $$ Here when asked to find…
madness
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Determine the solutions to $e^z - ie^{\pi} = 0$ which absolute value is smaller than $2\pi$

Given the equation $$e^z-ie^{\pi}=0 $$ solve the equation for which $|z|<2\pi$. My attempt Manipulating the equation $$e^z=ie^{\pi} $$ Let's take the natural log $$\ln(e^z)=\ln(i)\cdot \ln(e^{\pi}) $$ $$z = \frac{i\pi}{2}+\pi $$ So I guess this is…
Carl
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The value of $i^{-39}$

I have tried to solve the problem in the following way: $$i^{-39} = 1/i^{39} = i^{4\cdot 9+3} = 1/(-i) = -i$$ However the answer is supposed to be $i$ Did I do something wrong?
MsBonniePython
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