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de Moivre's formula


In previous post we have learned about complex numbers. Today, we will learn about the trigonometric form of a complex number and the famous de Moivre's formula.



Let us recall that the complex numbers are created by the introduction of a very special number -- the number $i$ -- with the property $$i^2 = -1.$$
Complex numbers have the form $$a + ib$$ where $a$ and $b$ are two real numbers. In previous post, we have learned some basic facts about complex numbers
  • Addition and Subtraction $$(a + i b) + (c + i d) = (a+c) + i (b+d) ,$$ $$(a + i b)- (c + i d) = (a-c) + i (b - d). $$
  • Multiplication $$(a + i b)(c + i d) = ac + i ad + i bc + i^2 bd = (ac - bd) + i (bc + ad ) .$$
  • Division we need to use the identity $$(a + i b)(a - ib ) = a^2 - i^2 b^2 = a^2 + b^2$$  and so $$\frac{c + i d}{a + i b} = \frac{(c + i d)(a - ib)}{(a + ib)(a - ib)} = \frac{(ac + bd) + i(ad - bc)}{a^2 + b^2} = \frac{ac + bd}{a^2 + b^2} + i \frac{ad - bc}{a^2 + b^2}.$$
  • Complex conjugate $$\overline{a + i b} = a - ib, ~~~~\overline{a- ib} = a + ib.$$
  • Absolute value $$|a + ib| = \sqrt{a^2 + b^2}.$$


Trigonometric form of a complex number

Today we will learn a very important fact about complex numbers, that is, any complex number $z$ can be written in a trigonometric form as follows $$z = r (\cos{\phi} + i ~\sin{\phi}),$$ where $r = |z|$.

Indeed, let $z = a + ib$, then $$r = |z| = \sqrt{a^2 + b^2},$$
we have $$\frac{z}{r} = \frac{a}{r} + i ~ \frac{b}{r}.$$

Since $$\left( \frac{a}{r} \right)^2 + \left( \frac{b}{r} \right)^2 = \frac{a^2 + b^2}{r^2} = 1,$$
there exists $\phi$ such that $$\frac{a}{r} = \cos{\phi}, ~~~~~~ \frac{b}{r} = \sin{\phi}.$$

Thus, $$\frac{z}{r} = \frac{a}{r} + i ~ \frac{b}{r} = \cos{\phi} + i ~ \sin{\phi}.$$

And we obtain the trigonometric form of the complex number $z$ $$z = r (\cos{\phi} + i ~\sin{\phi}).$$

In special case when $z = 0$ we can let $r=\alpha = 0$.


Multiplication in trigonometric form

Trigonometric form is very convenient when we do multiplication for complex numbers thanks to the following identity $$(\cos{\alpha} + i ~ \sin{\alpha})(\cos{\beta} + i ~\sin{\beta}) = \cos{(\alpha + \beta)} + i ~ \sin{(\alpha + \beta)} .$$

If we have two complex numbers $u$ and $v$, and express them in the trigonometric form $$u = r (\cos{\alpha} + i ~ \sin{\alpha}),$$ $$v = s (\cos{\beta} + i ~ \sin{\beta}),$$ then their product can be easily obtained as $$uv = rs (\cos{(\alpha + \beta) + i ~\sin{(\alpha + \beta)}}) .$$


Exponentiation and de Moivre's formula

Similar to multiplication, exponentiation is also easy when we write the complex number in trigonometric form. If $$u = r (\cos{\alpha} + i ~ \sin{\alpha})$$ then $$u^n = r^n (\cos{(n \alpha)} + i ~ \sin{(n \alpha)}).$$

The following identity is called the de Moivre's formula, this is an important formula in complex numbers $$(\cos{\alpha} + i ~ \sin{\alpha})^n =  \cos{(n \alpha)} + i ~ \sin{(n \alpha)}.$$ 

  

Now let us do some exercises. 


Problem 1: Solve the quadratic equation $$x^2 − 2 x + 4 =0$$ and write its complex roots in trigonometric form.

Solution: We have $$\Delta' = 1^2 - 4 = -3,$$ so the quadratic equation has complex roots $$1 \pm i~ \sqrt{3}.$$

To write these complex numbers in trigonometric form, we need to calculate their absolute value $$| 1 \pm i~ \sqrt{3} | = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.$$

Thus, $$1 \pm i~ \sqrt{3} = 2 ~\left( \frac{1}{2} \pm i ~\frac{\sqrt{3}}{2} \right) = 2 (\cos{\frac{\pi}{3}} \pm i ~ \sin{\frac{\pi}{3}}).$$


Problem 2: Solve the quadratic equation $$x^2 −  x + 1 =0$$ and write its complex roots in trigonometric form.

Solution: We have $$\Delta = 1^2 - 4 = -3,$$ so the quadratic equation has complex roots $$\frac{1 \pm i~ \sqrt{3}}{2}.$$

To write these complex numbers in trigonometric form, we need to calculate their absolute value $$\left| \frac{1 \pm i ~\sqrt{3}}{2} \right| = \sqrt{\left( \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2}\right)^2} = 1.$$

Thus, $$\frac{1 \pm i ~\sqrt{3}}{2} = \frac{1}{2} \pm i~ \frac{\sqrt{3}}{2} = \cos{\frac{\pi}{3}} \pm i ~ \sin{\frac{\pi}{3}}.$$


Problem 3: Solve the quadratic equation $$x^2 −  3 x + 3 =0$$ and write its complex roots in trigonometric form.

Solution: We have $$\Delta = 3^2 - 4 \times 3 = -3,$$ so the quadratic equation has complex roots $$\frac{3 \pm i~ \sqrt{3}}{2}.$$

To write these complex numbers in trigonometric form, we need to calculate their absolute value $$\left| \frac{3 \pm i ~\sqrt{3}}{2} \right| = \sqrt{\left( \frac{3}{2}\right)^2 + \left( \frac{\sqrt{3}}{2}\right)^2} = \sqrt{3}.$$

Thus, $$\frac{3 \pm i ~\sqrt{3}}{2} = \sqrt{3} \left( \frac{\sqrt{3}}{2} \pm i ~ \frac{1}{2}\right) = \sqrt{3} (\cos{\frac{\pi}{6}} \pm i ~ \sin{\frac{\pi}{6}}).$$


Problem 4: Calculate $(1 + i)^{2012}$ by two methods, using de Moivre's formula and using binomial identity, and deduce the following equality $${2012 \choose 0} - {2012 \choose 2} + {2012 \choose 4} - {2012 \choose 6} + \dots + {2012 \choose 2008} -  {2012 \choose 2010} + {2012 \choose 2012} = - 2^{1006}.$$

Solution: In the first method, we write $1+i$ in trigonometric form and then use de Moivre's formula to calculate the exponentiation.

First, we calculate the absolute value of $1+i$: $$|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}.$$

Thus, $$1 + i = \sqrt{2} \left( \frac{\sqrt{2}}{2} + i ~ \frac{\sqrt{2}}{2} \right) = \sqrt{2} ( \cos{\frac{\pi}{4}} + i ~ \sin{\frac{\pi}{4}}).$$

Using de Moivre's formula, we calculate the exponentiation $$(1 + i)^{2012} = (\sqrt{2})^{2012} ( \cos{\frac{2012 \pi}{4}} + i ~ \sin{\frac{2012 \pi}{4}}) = 2^{1006} (\cos{(503 \pi)} + i ~ \sin{(503 \pi)}) = - 2^{1006}.$$

On the other hand, by binomial identity, we have $$(1 + i)^{2012} = 1 + {2012 \choose 1} i + {2012 \choose 2} i^2 + {2012 \choose 3} i^3 + {2012 \choose 4} i^4 + {2012 \choose 5} i^5 + \dots + {2012 \choose 2011} i^{2011} + i^{2012}$$ $$= 1 + {2012 \choose 1} i - {2012 \choose 2} - {2012 \choose 3} i + {2012 \choose 4} + {2012 \choose 5} i + \dots - {2012 \choose 2011} i + 1$$

Comparing the real part of the above two results, we obtain $$1 - {2012 \choose 2} + {2012 \choose 4} - {2012 \choose 6} + \dots + {2012 \choose 2008} -  {2012 \choose 2010} + 1 = - 2^{1006}.$$



Let us stop here for now, see you again in the next post.




Homework.

1. Write these numbers in trigonometric form: $1 - i$, $3 + 3i$, $\sqrt{3} + 3i$, $3 - \sqrt{3} i$, $2$, $- 7 + 7i$, $3i$.

2. What is the absolute value of this complex number $\cos{\alpha} + i ~ \sin{\alpha}$ ?

3. Let $u = r (\cos{\alpha} + i ~ \sin{\alpha})$ and $v = s (\cos{\beta} + i ~ \sin{\beta})$, calculate $u/v$.

4. Calculate $(1 + i)^{2013}$ by two methods, the de Moivre's formula and the binomial identity, and then deduce some combinatorial equalities.

5. Write $x$ in trigonometric form and find all possible values of $x$ such that $x^4 = -1$.








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