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MATH
[1251]

Examples from Polar form

All examples from Polar form.

Back to Polar form

Convert between the two forms

For z=1+iz=1+i,

∣z∣=12+12=2.|z|=\sqrt{1^2+1^2}=\sqrt2.

The point lies in the first quadrant and tan⁡θ=1\tan\theta=1, so θ=π/4\theta=\pi/4 is a principal argument. Hence

1+i=2 eiπ/4.1+i=\sqrt2\,e^{i\pi/4}.

Conversely,

2e−3πi/4=2(cos⁡(−3π4)+isin⁡(−3π4))=−2−2 i.\begin{aligned} 2e^{-3\pi i/4} &=2\left(\cos\left(-\frac{3\pi}{4}\right) +i\sin\left(-\frac{3\pi}{4}\right)\right)\\ &=-\sqrt2-\sqrt2\,i. \end{aligned}

[routine]

  1. [1]:Polar form

Perform product and quotient calculations

Let

z1=2 e3πi/2,z2=πe3πi/2.z_1=\sqrt2\,e^{3\pi i/2}, \qquad z_2=\pi e^{3\pi i/2}.

Both factors point down the imaginary axis. Their arguments add to 3π3\pi, so the product points along the negative real axis:

z1z2=π2 e3πi=−π2.z_1z_2=\pi\sqrt2\,e^{3\pi i}= -\pi\sqrt2.

If instead z2=πe3πi/4z_2=\pi e^{3\pi i/4}, then

z1z2=2πei(3π/2−3π/4)=2πe3πi/4.\frac{z_1}{z_2} =\frac{\sqrt2}{\pi}e^{i(3\pi/2-3\pi/4)} =\frac{\sqrt2}{\pi}e^{3\pi i/4}.

Calculate a power

For z=2 e3πi/4z=\sqrt2\,e^{3\pi i/4},

z104=(2)104e104(3πi/4)=252e78πi=252.z^{104} =(\sqrt2)^{104}e^{104(3\pi i/4)} =2^{52}e^{78\pi i} =2^{52}.

Polar form handles the large exponent by treating the modulus and the argument separately.

[routine]

  1. [1]:De Moivre's formula

Find the eighth roots of unity

Since 1=e2πik1=e^{2\pi i k} for every integer kk, the eighth roots of unity are

zk=e2πik/8,k=0,1,…,7.z_k=e^{2\pi ik/8},\qquad k=0,1,\ldots,7.

They are the eight equally spaced points on the unit circle. For example,

z0=1,z1=22+22i,z2=i,z_0=1,\qquad z_1=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\qquad z_2=i,

and the remaining roots continue around the circle.

[routine]

  1. [1]:Find the nth roots

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