Usually, we represent the complex numbers, in the form of z = x+iy where ‘i’ the imaginary number. Following is a picture of \(w, z\), and \(wz\) that illustrates the action of the complex product. The word polar here comes from the fact that this process can be viewed as occurring with polar coordinates. The following figure shows the complex number z = 2 + 4j Polar and exponential form. Solution:7-5i is the rectangular form of a complex number. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Hence, it can be represented in a cartesian plane, as given below: Here, the horizontal axis denotes the real axis, and the vertical axis denotes the imaginary axis. Ms. Hernandez shows the proof of how to multiply complex number in polar form, and works through an example problem to see it all in action! Let 3+5i, and 7∠50° are the two complex numbers. Cos θ = Adjacent side of the angle θ/Hypotenuse, Also, sin θ = Opposite side of the angle θ/Hypotenuse. The following development uses trig.formulae you will meet in Topic 43. Roots of complex numbers in polar form. 4. How to solve this? Polar Form of a Complex Number. Now we write \(w\) and \(z\) in polar form. This is an advantage of using the polar form. The n distinct n-th roots of the complex number z = r( cos θ + i sin θ) can be found by substituting successively k = 0, 1, 2, ... , (n-1) in the formula. 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This is an advantage of using the polar form. So the polar form \(r(\cos(\theta) + i\sin(\theta))\) can also be written as \(re^{i\theta}\): \[re^{i\theta} = r(\cos(\theta) + i\sin(\theta))\]. The following applets demonstrate what is going on when we multiply and divide complex numbers. The following questions are meant to guide our study of the material in this section. Multiplication of Complex Numbers in Polar Form, Let \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) be complex numbers in polar form. To understand why this result it true in general, let \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) be complex numbers in polar form. \(\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) = \cos(\alpha - \beta)\), \(\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) = \sin(\alpha - \beta)\), \(\cos^{2}(\beta) + \sin^{2}(\beta) = 1\). 0. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers. Step 2. Multiply the numerator and denominator by the conjugate . We will use cosine and sine of sums of angles identities to find \(wz\): \[w = [r(\cos(\alpha) + i\sin(\alpha))][s(\cos(\beta) + i\sin(\beta))] = rs([\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)]) + i[\cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha)]\], We now use the cosine and sum identities and see that. To divide,we divide their moduli and subtract their arguments. Complex Numbers in Polar Form. Proof of the Rule for Dividing Complex Numbers in Polar Form. Division of Complex Numbers in Polar Form, Let \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) be complex numbers in polar form with \(z \neq 0\). What is the polar (trigonometric) form of a complex number? The conjugate of ( 7 + 4 i) is ( 7 − 4 i) . \[^* \space \theta = \dfrac{\pi}{2} \space if \space b > 0\] We now use the following identities with the last equation: Using these identities with the last equation for \(\dfrac{w}{z}\), we see that, \[\dfrac{w}{z} = \dfrac{r}{s}[\dfrac{\cos(\alpha - \beta) + i\sin(\alpha- \beta)}{1}].\]. Multiplication. Back to the division of complex numbers in polar form. Note that \(|w| = \sqrt{(-\dfrac{1}{2})^{2} + (\dfrac{\sqrt{3}}{2})^{2}} = 1\) and the argument of \(w\) satisfies \(\tan(\theta) = -\sqrt{3}\). When multiplying complex numbers in polar form, simply multiply the polar magnitudes of the complex numbers to determine the polar magnitude of the product, and add the angles of the complex numbers to determine the angle of the product: z = r z e i θ z . Here we have \(|wz| = 2\), and the argument of \(zw\) satisfies \(\tan(\theta) = -\dfrac{1}{\sqrt{3}}\). Let 2=−බ ∴=√−බ Just like how ℝ denotes the real number system, (the set of all real numbers) we use ℂ to denote the set of complex numbers… Determine real numbers \(a\) and \(b\) so that \(a + bi = 3(\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6}))\). This polar form is represented with the help of polar coordinates of real and imaginary numbers in the coordinate system. Let us learn here, in this article, how to derive the polar form of complex numbers. ⇒ z1z2 = r1eiθ1. Multiplication and division of complex numbers in polar form. Multiply & divide complex numbers in polar form Our mission is to provide a free, world-class education to anyone, anywhere. Figure \(\PageIndex{1}\): Trigonometric form of a complex number. Using our definition of the product of complex numbers we see that, \[wz = (\sqrt{3} + i)(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i) = -\sqrt{3} + i.\] Your email address will not be published. \[z = r{{\bf{e}}^{i\,\theta }}\] where \(\theta = \arg z\) and so we can see that, much like the polar form, there are an infinite number of possible exponential forms for a given complex number. Convert given two complex number division into polar form. We won’t go into the details, but only consider this as notation. N-th root of a number. How do we divide one complex number in polar form by a nonzero complex number in polar form? Using equation (1) and these identities, we see that, \[w = rs([\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)]) + i[\cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha)] = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\]. 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