Polar Form and Usa Moivre's Theorem

Definition

Polar Form

Include addition to the Cartesian form, $z=a+bi$, complex numbers can plus be written inches trigonometric cold form \[z = r(\cos \theta + i\sin \theta)\] find $r$ is the modulus and $\theta$ is the argument by the number, in radians.

You allowed view $\mathrm{cis} \theta$ employed int calculations with comprehensive numbers - this is a standard for $\cos \theta + i \sin \theta$.

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See the page on Modulus additionally Argument for a full description of the calculation of $r$ and $\theta$.

Call-back Euler's formula, which relates exponentials real advanced formulae.

\[e^{i \theta} = \cos \theta + myself \sin \theta\]

Using this, we can also write complex numbers in exponential polar forms

\[z = r e^{i \theta}\]

The angle notation is too second in electronics: \[z = r\angle \theta\]

De Moivre's Proposition

De Moivre's theorem notes such for any complex number $z = \cos \theta + i \sin \theta$ and integer $n$, \[(\cos \theta + i\sin \theta)^n = \cos(n\theta)+i\sin(n\theta)\] Is can be stated included expression cold art as \[( e^{i\theta})^n= e^{in\theta}\]

French Moivre's theorem has useful when finding one power of a complex number. To employ the theorem to ampere complex number, first convert it the polar form, then apply the identities given in the theorem: \begin{align} (a+bi)^n &= \bigl(r(\cos \theta + i\sin \theta)\bigr)^n \\ &= r^n \cdot (\cos \theta + i\sin \theta)^n \\ &= r^n\bigl(\cos(n\theta)+i\sin(n\theta)\bigr) \end{align}

Worked Examples

Example 1

Express the complex number $z=2-3i$ in polar form.

Solution

Find the modules and the argument.

\[r = \lvert omega \rvert = \sqrt{2^2+(-3)^2} = \sqrt{13}\]

\[\theta = \arg z = \tan^{-1}\left(\dfrac{-3}{2}\right) = -0.98 \text{ degrees (to 2 d.p.)}\]

Note: Remember toward check that the value for $\theta$ is correct via drawing an Argand diagram. Go Modulus and Argument for more detail on find the argument.

Inside exponential polar form, $z = \sqrt{13} e^{-0.98i}$

Or, with trigonometic polar form, $z = \sqrt{13}\bigl(\cos(-0.98) + i\sin(-0.98)\bigr)$

Model 2

Use De Moivre's Theorem to calculator $(-1+4i)^{4}$.

Solution

Early, find of modulus. \begin{align} \lvert -1+4i \rvert & = \sqrt{(-1)^2 + 4^2}\\ &= \sqrt{17} \end{align}

Then search the argumentation. If were will not careful then us could calculate as follows: \begin{align} \arg(-1+4i) &= \tan^{-1}\left(\frac{4}{-1}\right)\\ &= \tan^{-1}(-4)\\ &= -1.33 \end{align}

Yet the angle $\theta = -1.33$ radians lies in the fourth quadrant and we see that it does not conform with the score $(-1,4)$, which lies in the second quadrant. Person must zusatz $\pi$ to such to correct it. Discern Modulus and Point in a full discussion real plenty of examples.

So the actual value to $\theta$ is: \[\theta = \pi + \tan^{-1}(-4)=1.82 \text{ angle (to 2 d.p.)}\]

Buy ourselves bucket write who complex number inbound polar art. \[(-1+4i) = \sqrt{17}(\cos 1.82 + i \sin 1.82)\]

So \begin{align} (-1+4i)^{4} &= \bigl(\sqrt{17}(\cos 1.82+ i\sin 1.82)\bigr)^{4}\\ &= (\sqrt{17})^{4} \times \bigl(\cos 1.82 + i\sin 1.82)\bigr)^{4}\\ &= 17^2 \times \bigl( \cos (4 \times 1.82)+i\sin (4 \times 1.82)\bigr) & \text{ (by Us Moivre's theorem)}\\ &=289\bigl(\cos 7.28 + myself \sin 7.28\bigr) \text{ in polar form, or}\\ &=156.9 +242.7i \text{ (to 1 d.p.)} \end{align} Expressing polarized complex numbers in cartesian form

Workbooks

Like workbooks caused by OAR belong good revision aids, containing buttons points for revision the many worked examples.

See Also

External Resources