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This can be interpreted geometrically that multiplying by a complex number causes rotation in the complex plane.
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We can use this proprety to easily compute powers of complex numbers, which is sometimes called **DeMoivre's theorem**:
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Note that when we take the conjugate of a number $z = r e^{i\theta}$ we get $\overline{z} = r^{-i\theta}.$ This is a number with the same magnitude as $z$ but with the opposite angle. On the unit circle, where all numbers have the same magnitude, we have $z \overline{z} = 1$ because multiplying $z$ by its conjugate rotates it exactly how much it was rotated, but in the opposite direction.
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We can use this property to easily compute powers of complex numbers, which is sometimes called **DeMoivre's theorem**:
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:::theorem "de Moivre's Theorem" {label: de-moivres-}
@@ -139,9 +141,9 @@ $$ X = x^2 - y^2, \quad Y = 2xy $$
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Now we have two equations and two unknowns, and when they are solved, we will have the two square roots of $Z$.
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By definition, the **principal square root** of $Z$ is the square root with the nonnegative real part.
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By definition, the **principal square root** of $Z$ is the square root with the non-negative real part.
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## Complex Quadratrics
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## Complex Quadratics
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We can use the quadratic formula to solve quadratics in a complex variable with complex coefficients, that is where
These branches are analytic on any domain that does not contain $z = 0$ (the branch point) or points on the branch cut (the half-line through $z=0$ making an angle of $\phi$ radians ith the positive real axis.)
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We can define branch cuts that aren't straight lines too. For example, branch points of the function $\Log{\left \[ f(z)\right \]}$ are the zeroes of $f(z)$ and branch cuts are where $f(z)$ is real and negative.
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We can define branch cuts that aren't straight lines too. For example, branch points of the function $\Log{ \left [ f(z) \right ] }$ are the zeroes of $f(z)$ and branch cuts are where $f(z)$ is real and negative.
So, our radius of convergence here is $1/L.$ Note that if we had something like $w = (z - z_0)^{mn},$ then our radius of convergence would be $(1/L)^{1/m}.$
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So, our radius of convergence for our $w$ series here is $1/L.$ We have to bring $z$ back, so for example, if we had something like $w = (z - z_0)^{mn},$ then our radius of convergence would be $(1/L)^{1/m}.$ We get that by
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$$ |w| < R \implies |z^m| < R \implies |z| < R^{1/m}. $$
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Similarly with the root test, we end up with something like
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