Roots of a complex number

To solve the equation [math]z^n=a+bi[/math] for [math]a,b\in \mathbb R[/math] and [math]n\in \mathbb N[/math]
Change the values of a, b, and n.

The function f(z)=1/z (Part 1)

Mapping lines and circles with the function [math]f(z)=\frac{1}{z}[/math]
The function f(z)=1/z (Part 1)

Riemann Surface: z^(1/2)

GGB script
Re1(x, y) = exp(1 / 2 * log( sqrt( x*x + y*y ) ) ) * cos(1 / 2 * arctan2( y, x ))[br]Im1(x, y) = exp(1 / 2 * log( sqrt( x*x + y*y ) ) ) * sin(1 / 2 * arctan2( y, x ))[br][br]t = Slider(0, 1, 0.1, 1, 100, false, true, false, false)[br][br]HRe1(x, y) = 0 * (1-t) + Re1(x, y) * t[br]HIm1(x, y) = 0 * (1-t) + Im1(x, y) * t[br][br]Re2(x, y) = exp(1 / 2 * log( sqrt( x * x + y*y ) ) ) * cos(1 / 2 * (arctan2( y, x ) + 2 pi))[br]Im2(x, y) = exp(1 / 2 * log( sqrt( x * x + y*y ) ) ) * sin(1 / 2 * (arctan2( y, x )+2 pi))[br][br]HRe2(x, y) = 0 * (1-t) + Re2(x, y) * t[br]HIm2(x, y) = 0 * (1-t) + Im2(x, y) * t[br][br]RSRe = Surface(u*cos(v), u*sin(v), HRe1(u*cos(v), u*sin(v)), u, 0, 2, v, -pi, pi-pi/200)[br]RSIm = Surface(u*cos(v), u*sin(v), HIm1(u*cos(v), u*sin(v)), u,0, 2, v, -pi, pi-pi/200)[br]RSReN = Surface(u*cos(v), u*sin(v), HRe2(u*cos(v), u*sin(v)), u, 0, 2, v, -pi, pi-pi/200)[br]RSImN = Surface(u*cos(v), u*sin(v), HIm2(u*cos(v), u*sin(v)), u, 0, 2, v, -pi, pi-pi/200)

Riemann sphere

The extended complex plane can be mapped onto the surface of a sphere whose south [br]pole corresponds to the origin and whose north pole to the point [math]\infty[/math].
Move the slider x and y to move the point [math]z=x+iy[/math]

Series (Part I)

An infinite series [math]\sum_{n=1}^{\infty}z_n[/math] of complex numbers converges to the sum [math]S[/math] if the sequence [math]S_N=\sum_{n=1}^{N}z_n[/math] of partial sums converges to [math]S[/math]. We then write [math]\sum_{n=1}^{\infty}z_n=S[/math]. For example, [math]\sum_{n=1}^{\infty}z^n=\frac{1}{1-z}[/math].[br][br]Drag the point [math]z[/math] around to see what happens to the partial sums when [math]z[/math] is inside or out of the open ball [math]|z|=1[/math].
Series (Part I)

Hydrodynamics: Streamlines and velocity potentials

If we have a (steady-state) incompressible, nonviscous fluid, we are interested in finding its velocity field [math]V(x,y).[/math] [br]If [math]F=\phi+i\psi[/math] is the complex potential, the functions [math]\phi[/math] and [math]\psi[/math] are the [b]velocity potential[/b] and [b]stream[/b] functions, respectively. In this case [math]V=\text{grad}\,\phi[/math].
Hydrodynamics: Streamlines and velocity potentials

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