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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Demonstration with complex number

How can I show that $$\overline{e^{i\theta}}=e^{-i\theta}$$ I know that if z is a complex number so $$\overline{e^z}=e^\bar{z}$$ But i don't understand how to show this result.
Roland
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Complex number-argands diagram

The complex numbers, $z$ and $w$ satisfy the inequalities $|z-3-2i|\le2$ and $|w-7-5i|\le 1$. Find the least possible value of $|z-w|$. Thats my work till now $$|z-3-2i| = |z-(3+2i)| |z|-|3+2i| |z|-|3+2i| \le 2 |z|^2-|3+2i|^2 \le 2^2…
Abmon98
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Spotting a complex number

Can anyone spot a complex number $z\in\mathbb C$ such that for $a,b\in \mathbb C$ and $f(w)=\dfrac{w-z}{w-\overline{z}}$, we have $\dfrac{f(a)-f(b)}{1-\overline{f(a)}f(b)}=\dfrac{b-a}{b-\overline{a}}$. I have been staring at this for some time…
Aida
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What does $\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3} $ imply?

I'm having trouble understanding what the following equality implies. $$\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$ I suspect that this means that the points form the vertices of an equilateral triangle in the complex plane, but can't prove…
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Square Rooting Back To Real Dimension

As we all know, square rooting -1 (a real number) opens up the "imaginary" dimension (defined by the presence of iota). We can return from the imaginary dimension back to the real dimension by reversing this process (i.e by taking a square of…
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Complex numbers equation: $z^4 = -16$

Probably dumb question, but I would like to ask it anyway. I was to solve this equation: $z^4 = -16$ At first glance, the way to solve it would be (as any other equation): $z^2 = \sqrt{-1} * 4 \lor z^2 = -\sqrt{-1} * 4$ $z^2 = 4i \lor z^2 = -4i$ $z…
Griwes
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How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$

I have these problems : How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ For some reason this is incorrect I'll be glad to understand why, This is what I done : I used this formula :…
JaVaPG
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Factoring polynomial with complex coefficients

Given the equation $z^2+4iz-13=0$, solve for $z$ without the quadratic formula. In real numbers set, when I find this kind of equations I usually complete the perfect square trinomial.In this case: $(z^2+4iz-4)-13+4=0$ $(z+2i)^2-9=0$ I chosen $-4$…
user24047
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$\cosh(iz) -\cosh(z)=0$

$\cosh(iz) -\cosh(z)=0$ Apparently $iz=z$ and $z=0$ is a solution. How do I proceed next? Do I need to convert $\cosh$ into $\exp$ form? I tried that I get $e$ to complex and real power and I don't know how to continue. Please help!
John Lennon
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Problem involving complex conjugate

I have the equation: $$3z-\bar{z}=2-3i$$ First I write this as: $$3(x+yi)-(x-yi) = 2-3i$$ $$3x-3yi-x+yi = 2-3i$$ $$2x+4yi = 2-3i$$ Now the following must be true: $$2x = 2\quad\mbox{and}\quad 4y = -3$$ So $x = 1$, and…
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How do I solve $x^4+44=0$ according to de Moivre?

How do I solve $x^4+44=0$ according to de Moivre? I tried to use the formula, but I got roots that are not beautiful numbers. What should the complex roots for this equation be?
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Linear Algebra - Complex equation

I have this problem : $$x^2=i$$ The args = $\pi/2$. $r = |z| = \sqrt{0^2+i^2}=\sqrt{i^2}=i$ for $$z_0=i((\cos (\pi/2)/2)+isin(\pi/2)/2)) = i(\frac{\sqrt{2}}{2})+i\frac{\sqrt{2}}{2})=i\frac{\sqrt{2}}{2}+i^2(\frac{\sqrt{2}}{2})$$ For some reason I…
JaVaPG
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Number of roots of a complex equation

The question asks me to show that "in general," $$z \overline{z} + az + \overline{b}z + c = 0$$ has two solutions, and asks when it has more than two. Its obvious that if $a + b = 0$ and $c$ is negative, we get an equation of a circle. Now, the two…
Geano
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Finding the cube roots of $-i$.

As the title suggests, I am having trouble finding the cube roots of $-i$. I looked at the similar post of finding the cube roots of just $i$ but I am having some confusion. Complex numbers are of the form $z=a+bi$. $$-i= 0+(-1)i$$ Giving a modulus…
Vincent
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Solve for $z$ in $\sin(iz) = 4i$

When solving this equation, I reach a point where $$e^{2z} =\frac{1-8i}{65}$$ Now I'm pretty sure this is correct since this is what the memo has as well. But what I would like to know is why I cannot simply let $z= a+bi$ and then solve as…
dan
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