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6.3 Trigonometry and Complex Numbers

Trigonometric form of complex numbers

Let \(z = a + b i\) denote a complex number with real part \(a\) and imaginary part \(bi\).

Then the absolute value or modulus of \(z\), denoted \(| z |\), is the distance from \(z\) to zero in the complex plane and is computed by the formula \(|z|=\sqrt{a^2+b^2}\).

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The line containing \(0\) and \(z\) makes an angle \(t\) with the real axis satisfying \(\tan t=\frac{b}{a}\). The angle \(t\) is called the argument of \(z\) and may be styled \(t=\arg z\).

Furthermore,

\(\cos t = \dfrac{a}{| z |}\) and \(\sin t = \dfrac{b}{| z |}\).

Thus \(a = | z | \cos t\) and \(b = | z | \sin t\). So \(z = | z | \cos t + \left(| z | \sin t\right) i\).

So \(z = | z | (\cos t + i \sin t)\).

This is the trigonometric form of a complex number.

Complex product in trigonometric form

Suppose \(z_1\) and \(z_2\) are complex numbers. Then their product is

\(\begin{eqnarray*} z_1 \cdot z_2 &=& | z_1 | ( \cos t_1 + i \sin t_1 )\cdot | z_2 |( \cos t_2 + i \sin t_2 )\\ &=& | z_1 | | z_2 | ( \cos t_1 \cos t_2 - \sin t_1 \sin t_2 ) + i ( \cos t_1 \sin t_2 + \sin t_1 \cos t_2)\\ &=& | z_1 | | z_2 | \left(\cos ( t_1 + t_2 ) + i\, \sin ( t_1 + t_2 )\right) \end{eqnarray*}\)

Thus,

\[ z_1 \cdot z_2 = | z_1 | | z_2 | \left[ \cos ( t_1 + t_2 ) + i\, \sin ( t_1 + t_2 ) \right] \]

Exercise 6.3.1

Let \(z = 3 + 3 i\).

Find \( |z|\), the modulus of \(z\).

Find an angle \(t=\arg z\) in radians satifying \(z = | z | ( \cos t + i \sin t )\).

Write \(z\) in trigonometric form.

See Solution

Exercise 6.3.2

Show that \(z^2 = | z |^2 ( \cos 2t + i\,\sin 2t )\).

See Solution

DeMoivre's Theorem

The result in Exercise 6.3.2 can be extended to the general result, called DeMoivre's Theorem:

If \(z = | z | ( \cos t + i\, \sin t )\), then \(z^n = | z |^n [ \cos ( n t ) + i\, \sin ( n t ) ]\).

This means that the modulus of \(z^n\) is \(|z|^n\) and \(\arg z^n=n\arg z\).

Example: Find \(( -1 + 2 i )^5\)

Let \(z = -1 + 2 i\). Then \(\mod z^5=|z|^5=\left(\sqrt{5}\right)^5\approx55.9\).

Next, we must find the angle \(t=\arg(-1+2 i)\).

Since \(z\) lies in the second quadrant of the complex plane, we are looking for an angle \(t\) in the second quadrant satisfying \(\tan t = \frac{b}{a} = \frac{2}{-1}=-2\). \(\arctan ( -2 )\) will not do, since it is in quadrant IV. We want \(\pi + \arctan ( -2 )\), which is approximately \(2.03444\). [The calculator must be in radian mode.]

Thus, by DeMoivre's Theorem \(( -1 + 2 i )^5\) is approximately equal to

\(\begin{eqnarray*} 55.9 [ \cos 5(2.0344) + i \sin 5(2.0344) ]&=& 55.9(\cos10.1720 + i\sin10.1720)\\ &=& 55.9 ( -0.7336 - 0.6796 i )\\ &=& - 41.0082 - 37.9896 i. \end{eqnarray*}\)

Since the real and imaginary parts of \(( -1 + 2 i )^5\) must be integers, the exact answer is

\(( -1 + 2 i )^5 = - 41 - 38 i\).

Exercise 6.3.3

Find \(( 1 + i )^7\).

See Solution

Roots of Complex Numbers

The inverse of finding powers of complex numbers is finding roots of complex numbers.

A complex number has two square roots, three cube roots, four fourth roots, etc. Generally, a complex number has \(n\) nth roots. All \(n\) of the \(n\)th roots of a complex number \(z\) are evenly spaced around a circle centered at \(0\) and having a radius equal to the \(n\)th root of the modulus of \(z\). Furthermore, if \(z = | z | ( \cos t + i \sin t )\), then one of the \(n\) \(n\)th roots of \(z\) is \(w_0=| z |^{1 / n} \left[ \cos \left( \frac{t}{n} \right) + i \sin \left( \frac{t}{n} \right) \right]\). Since the \(\)n \(n\)th roots are evenly spaced around a circle, they are separated by an angle of \(\frac{2\pi}{n}\). So, once we find the root \(| z |^{1 / n} \left[ \cos \left( \frac{t}{n} \right) + i \sin \left( \frac{t}{n} \right) \right]\), we find the remaining \(n - 1\) roots by adding multiples of \(\frac{2\pi}{n}\) to \(\frac{t}{n}\).

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Example: Find the cube roots of \(1 + i\).

First, write \(1 + i\) in trigonometric form: \(z=1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\,\sin\frac{\pi}{4}\right)\)

Then one root is \(w_0=\sqrt[6]{2}\left(\cos\frac{\pi}{12}+i\,\sin\frac{\pi}{12}\right)\).

To find the arguments of the two remaining cube roots, we add multiples of \(\frac{2\pi}{3}=\frac{8\pi}{12}\) to \(\frac{\pi}{12}\) to get \(\frac{9\pi}{12}=\frac{3\pi}{4}\) and \(\frac{17\pi}{12}\).

Thus the three cube roots of \(1 + i\) are

\(w_0=\sqrt[6]{2}\left(\cos\frac{\pi}{12}+i\,\sin\frac{\pi}{12}\right)\)

\(w_1=\sqrt[6]{2}\left(\cos\frac{3\pi}{4}+i\,\sin\frac{3\pi}{4}\right)\)

\(w_2=\sqrt[6]{2}\left(\cos\frac{17\pi}{12}+i\,\sin\frac{17\pi}{12}\right)\)

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Exercise 6.3.4

Find the four fourth roots of \(-16 i\).

See Solution