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MATH
[1251]

Theorems from Polar form

All theorems from Polar form.

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Modulus laws

For z,z1,z2∈Cz,z_1,z_2\in\mathbb C with z2≠0z_2\ne0,

zz‾=∣z∣2,∣z1z2∣=∣z1∣∣z2∣,∣z1z2∣=∣z1∣∣z2∣,z\overline z=|z|^2,\qquad |z_1z_2|=|z_1||z_2|,\qquad \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|},∣Re⁡(z)∣≤∣z∣,∣Im⁡(z)∣≤∣z∣,∣z1+z2∣≤∣z1∣+∣z2∣,|\operatorname{Re}(z)|\le|z|, \qquad |\operatorname{Im}(z)|\le|z|, \qquad |z_1+z_2|\le|z_1|+|z_2|,∣z1−z2∣≥∣∣z1∣−∣z2∣∣,∣∑j=1Nzj∣≤∑j=1N∣zj∣(N≥1).|z_1-z_2|\ge\bigl||z_1|-|z_2|\bigr|, \qquad \left|\sum_{j=1}^N z_j\right| \le\sum_{j=1}^N|z_j|\quad(N\ge1).

The inequality for z1+z2z_1+z_2 is the triangle inequality in the Argand plane; the finite-sum version follows by applying it repeatedly. The reverse triangle inequality follows from

∣z1∣=∣(z1−z2)+z2∣≤∣z1−z2∣+∣z2∣|z_1|=|(z_1-z_2)+z_2|\le |z_1-z_2|+|z_2|

and the same argument with z1z_1 and z2z_2 interchanged.

De Moivre's formula

For θ∈R\theta\in\mathbb R and n∈Zn\in\mathbb Z,

(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ).\left(\cos\theta+i\sin\theta\right)^n =\cos(n\theta)+i\sin(n\theta).

More generally, if z=reiθz=re^{i\theta} with r>0r>0, then

zn=rneinθ.z^n=r^ne^{in\theta}.

For positive integers, use induction. The case n=1n=1 is immediate and if the formula holds at n=kn=k, multiplying by reiθre^{i\theta} gives

zk+1=rkeikθreiθ=rk+1ei(k+1)θ.z^{k+1}=r^ke^{ik\theta}re^{i\theta}=r^{k+1}e^{i(k+1)\theta}.

The zero case is z0=1z^0=1 and the negative-integer case follows from the multiplicative inverse. The condition r>0r>0 is essential for negative powers, because 00 has no multiplicative inverse.

Find the nth roots

If z0≠0z_0\ne0, nn is a positive integer and z0=r0eiθ0z_0=r_0e^{i\theta_0}, its distinct nnth roots are

zk=r0nexp⁡(iθ0+2πkn),k=0,1,…,n−1.z_k=\sqrt[n]{r_0}\exp\left(i\frac{\theta_0+2\pi k}{n}\right), \qquad k=0,1,\ldots,n-1.

They lie on a circle of radius r0n\sqrt[n]{r_0} centred at the origin and are equally spaced by an angle of 2π/n2\pi/n. A non-zero complex number has exactly nn distinct nnth roots.

To derive the formula, write z=ρeiϕz=\rho e^{i\phi}. If zn=z0z^n=z_0, equality of moduli gives ρn=r0\rho^n=r_0, while equality of arguments gives nϕ=θ0+2πkn\phi=\theta_0+2\pi k. Taking k=0,…,n−1k=0,\ldots,n-1 produces all distinct possibilities: two listed angles differ by less than 2π2\pi, so they cannot describe the same point. Replacing kk by k+nk+n adds 2π2\pi to the root's angle, so any other integer kk repeats one of them. If z0=0z_0=0, the only nnth root is z=0z=0.

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