Since \(z\) is in the first quadrant, we know that \(\theta = \dfrac{\pi}{6}\) and the polar form of \(z\) is \[z = 2[\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6})]\], We can also find the polar form of the complex product \(wz\). Complex Number Division Formula, what is a complex number, roots of complex numbers, magnitude of complex number, operations with complex numbers. Example: Find the polar form of complex number 7-5i. We know, the modulus or absolute value of the complex number is given by: To find the argument of a complex number, we need to check the condition first, such as: Here x>0, therefore, we will use the formula. \(\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\) and \(\sin(\alpha + \beta) = \cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha)\). We know the magnitude and argument of \(wz\), so the polar form of \(wz\) is, \[wz = 6[\cos(\dfrac{17\pi}{12}) + \sin(\dfrac{17\pi}{12})]\]. You da real mvps! \(\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) = \cos(\alpha - \beta)\), \(\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) = \sin(\alpha - \beta)\), \(\cos^{2}(\beta) + \sin^{2}(\beta) = 1\). 4. (Argument of the complex number in complex plane) 1. Following is a picture of \(w, z\), and \(wz\) that illustrates the action of the complex product. 1. The parameters \(r\) and \(\theta\) are the parameters of the polar form. by M. Bourne. $1 per month helps!! Proof of the Rule for Dividing 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. The terminal side of an angle of \(\dfrac{17\pi}{12} = \pi + \dfrac{5\pi}{12}\) radians is in the third quadrant. \]. In polar form, the multiplying and dividing of complex numbers is made easier once the formulae have been developed. \[z = r(\cos(\theta) + i\sin(\theta)). \[^* \space \theta = \dfrac{\pi}{2} \space if \space b > 0\] Let \(w = 3[\cos(\dfrac{5\pi}{3}) + i\sin(\dfrac{5\pi}{3})]\) and \(z = 2[\cos(-\dfrac{\pi}{4}) + i\sin(-\dfrac{\pi}{4})]\). When performing addition and subtraction of complex numbers, use rectangular form. Missed the LibreFest? So \[3(\cos(\dfrac{\pi}{6} + i\sin(\dfrac{\pi}{6})) = 3(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i) = \dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i\]. To better understand the product of complex numbers, we first investigate the trigonometric (or polar) form of a complex number. The argument of \(w\) is \(\dfrac{5\pi}{3}\) and the argument of \(z\) is \(-\dfrac{\pi}{4}\), we see that the argument of \(\dfrac{w}{z}\) is, \[\dfrac{5\pi}{3} - (-\dfrac{\pi}{4}) = \dfrac{20\pi + 3\pi}{12} = \dfrac{23\pi}{12}\]. 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 \(z \neq 0\). Determine the polar form of \(|\dfrac{w}{z}|\). Also, \(|z| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = 2\) and the argument of \(z\) satisfies \(\tan(\theta) = \dfrac{1}{\sqrt{3}}\). z = r z e i θ z. z = r_z e^{i \theta_z}. 4. A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. What is the complex conjugate of a complex number? In which quadrant is \(|\dfrac{w}{z}|\)? The argument of \(w\) is \(\dfrac{5\pi}{3}\) and the argument of \(z\) is \(-\dfrac{\pi}{4}\), we see that the argument of \(wz\) is \[\dfrac{5\pi}{3} - \dfrac{\pi}{4} = \dfrac{20\pi - 3\pi}{12} = \dfrac{17\pi}{12}\]. Let z 1 = r 1 cis θ 1 and z 2 = r 2 cis θ 2 be any two complex numbers. In this section, we studied the following important concepts and ideas: If \(z = a + bi\) is a complex number, then we can plot \(z\) in the plane. Derivation What is the polar (trigonometric) form of a complex number? Roots of complex numbers in polar form. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 5.2: The Trigonometric Form of a Complex Number, [ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:tsundstrom", "modulus (complex number)", "norm (complex number)" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FPrecalculus%2FBook%253A_Trigonometry_(Sundstrom_and_Schlicker)%2F05%253A_Complex_Numbers_and_Polar_Coordinates%2F5.02%253A_The_Trigonometric_Form_of_a_Complex_Number, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 5.3: DeMoivre’s Theorem and Powers of Complex Numbers, ScholarWorks @Grand Valley State University, Products of Complex Numbers in Polar Form, Quotients of Complex Numbers in Polar Form, Proof of the Rule for Dividing Complex Numbers in Polar Form. Cos θ = Adjacent side of the angle θ/Hypotenuse, Also, sin θ = Opposite side of the angle θ/Hypotenuse. To convert into polar form modulus and argument of the given complex number, i.e. Multiply & divide complex numbers in polar form Our mission is to provide a free, world-class education to anyone, anywhere. Watch the recordings here on Youtube! This is an advantage of using the polar form. We won’t go into the details, but only consider this as notation. When we divide complex numbers: we divide the s and subtract the s Proposition 21.9. Back to the division of complex numbers in polar form. z 1 z 2 = r 1 cis θ 1 . To prove the quotation theorem mentioned above, all we have to prove is that z1 z2 in the form we presented, multiplied by z2, produces z1. Use right triangle trigonometry to write \(a\) and \(b\) in terms of \(r\) and \(\theta\). In general, we have the following important result about the product of two complex numbers. Convert given two complex number division into polar form. [See more on Vectors in 2-Dimensions].. We have met a similar concept to "polar form" before, in Polar Coordinates, part of the analytical geometry section. Division of complex numbers means doing the mathematical operation of division on complex numbers. 3. 3. This is the polar form of a complex number. Using our definition of the product of complex numbers we see that, \[wz = (\sqrt{3} + i)(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i) = -\sqrt{3} + i.\] Let z1 =r1eiθ1 and z2 =r2eiθ2 z 1 = r 1 e i θ 1 a n d z 2 = r 2 e i θ 2. Euler's formula for complex numbers states that if z z z is a complex number with absolute value r z r_z r z and argument θ z \theta_z θ z , then . r 2 cis θ 2 = r 1 r 2 (cis θ 1 . Now, we need to add these two numbers and represent in the polar form again. Also, \(|z| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}\) and the argument of \(z\) is \(\arctan(\dfrac{-1}{1}) = -\dfrac{\pi}{4}\). Note that \(|w| = \sqrt{4^{2} + (4\sqrt{3})^{2}} = 4\sqrt{4} = 8\) and the argument of \(w\) is \(\arctan(\dfrac{4\sqrt{3}}{4}) = \arctan\sqrt{3} = \dfrac{\pi}{3}\). Free Complex Number Calculator for division, multiplication, Addition, and Subtraction Note that \(|w| = \sqrt{(-\dfrac{1}{2})^{2} + (\dfrac{\sqrt{3}}{2})^{2}} = 1\) and the argument of \(w\) satisfies \(\tan(\theta) = -\sqrt{3}\). by M. Bourne. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . The equation of polar form of a complex number z = x+iy is: Let us see some examples of conversion of the rectangular form of complex numbers into polar form. 0. 1. How to algebraically calculate exact value of a trig function applied to any non-transcendental angle? So \[z = \sqrt{2}(\cos(-\dfrac{\pi}{4}) + \sin(-\dfrac{\pi}{4})) = \sqrt{2}(\cos(\dfrac{\pi}{4}) - \sin(\dfrac{\pi}{4})\], 2. Writing a Complex Number in Polar Form Plot in the complex plane.Then write in polar form. When we write \(z\) in the form given in Equation \(\PageIndex{1}\):, we say that \(z\) is written in trigonometric form (or polar form). Draw a picture of \(w\), \(z\), and \(|\dfrac{w}{z}|\) that illustrates the action of the complex product. 5 + 2 i 7 + 4 i. N-th root of a number. Multiplication. Multiplication and division in polar form Introduction When two complex numbers are given in polar form it is particularly simple to multiply and divide them. Let and be two complex numbers in polar form. a =-2 b =-2. Multiplication of complex numbers is more complicated than addition of complex numbers. Khan Academy is a 501(c)(3) nonprofit organization. Recall that \(\cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2}\) and \(\sin(\dfrac{\pi}{6}) = \dfrac{1}{2}\). Complex Numbers in Polar Form. if z 1 = r 1∠θ 1 and z 2 = r 2∠θ 2 then z 1z 2 = r 1r 2∠(θ 1 + θ 2), z 1 z 2 = r 1 r 2 ∠(θ 1 −θ 2) Note that to multiply the two numbers we multiply their moduli and add their arguments. There is an alternate representation that you will often see for the polar form of a complex number using a complex exponential. This turns out to be true in general. Let n be a positive integer. Hence, the polar form of 7-5i is represented by: Suppose we have two complex numbers, one in a rectangular form and one in polar form. 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. Then the polar form of the complex quotient \(\dfrac{w}{z}\) is given by \[\dfrac{w}{z} = \dfrac{r}{s}(\cos(\alpha - \beta) + i\sin(\alpha - \beta)).\]. We now use the following identities with the last equation: Using these identities with the last equation for \(\dfrac{w}{z}\), we see that, \[\dfrac{w}{z} = \dfrac{r}{s}[\dfrac{\cos(\alpha - \beta) + i\sin(\alpha- \beta)}{1}].\]. There is a similar method to divide one complex number in polar form by another complex number in polar form. The angle \(\theta\) is called the argument of the argument of the complex number \(z\) and the real number \(r\) is the modulus or norm of \(z\). Multiplication and Division of Complex Numbers in Polar Form Key Questions. This video gives the formula for multiplication and division of two complex numbers that are in polar form… 4. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. For complex numbers with modulo #1#, geometrically, multiplication is a rotation of a vector representing the first complex number counterclockwise by the angle of the second number. The result of Example \(\PageIndex{1}\) is no coincidence, as we will show. Explain. z = r z e i θ z . When we compare the polar forms of \(w, z\), and \(wz\) we might notice that \(|wz| = |w||z|\) and that the argument of \(zw\) is \(\dfrac{2\pi}{3} + \dfrac{\pi}{6}\) or the sum of the arguments of \(w\) and \(z\). Multiplication and division of complex numbers in polar form. This states that to multiply two complex numbers in polar form, we multiply their norms and add their arguments, and to divide two complex numbers, we divide their norms and subtract their arguments. Let's divide the following 2 complex numbers. Using equation (1) and these identities, we see that, \[w = rs([\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)]) + i[\cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha)] = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\]. \[z = r{{\bf{e}}^{i\,\theta }}\] 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. z =-2 - 2i z = a + bi, This polar form is represented with the help of polar coordinates of real and imaginary numbers in the coordinate system. Since \(|w| = 3\) and \(|z| = 2\), we see that, 2. Let us learn here, in this article, how to derive the polar form of complex numbers. Usually, we represent the complex numbers, in the form of z = x+iy where ‘i’ the imaginary number. The proof of this is best approached using the (Maclaurin) power series expansion and is left to the interested reader. Hence. To better understand the product of complex numbers, we first investigate the trigonometric (or polar) form of a complex number. Products and Quotients of Complex Numbers. So, \[\dfrac{w}{z} = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha) + i\sin(\alpha))}{(\cos(\beta) + i\sin(\beta)} \right ] = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha) + i\sin(\alpha))}{(\cos(\beta) + i\sin(\beta)} \cdot \dfrac{(\cos(\beta) - i\sin(\beta))}{(\cos(\beta) - i\sin(\beta)} \right ] = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)) + i(\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)}{\cos^{2}(\beta) + \sin^{2}(\beta)} \right ]\]. If \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) are complex numbers in polar form, then the polar form of the complex product \(wz\) is given by, \[wz = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\] and \(z \neq 0\), the polar form of the complex quotient \(\dfrac{w}{z}\) is, \[\dfrac{w}{z} = \dfrac{r}{s}(\cos(\alpha - \beta) + i\sin(\alpha - \beta)),\]. Then the polar form of the complex product \(wz\) is given by, \[wz = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\]. With Euler’s formula we can rewrite the polar form of a complex number into its exponential form as follows. The following questions are meant to guide our study of the material in this section. How do we multiply two complex numbers in polar form? \[^* \space \theta = -\dfrac{\pi}{2} \space if \space b < 0\], 1. To find \(\theta\), we have to consider cases. We illustrate with an example. Required fields are marked *. Figure \(\PageIndex{2}\): A Geometric Interpretation of Multiplication of Complex Numbers. Let us consider (x, y) are the coordinates of complex numbers x+iy. How to solve this? Multipling and dividing complex numbers in rectangular form was covered in topic 36. We can think of complex numbers as vectors, as in our earlier example. Multiply the numerator and denominator by the conjugate . 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. First, we will convert 7∠50° into a rectangular form. Polar Form of a Complex Number. Now we write \(w\) and \(z\) in polar form. This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers. Complex numbers are often denoted by z. Determine the polar form of the complex numbers \(w = 4 + 4\sqrt{3}i\) and \(z = 1 - i\). divide them. Therefore, the required complex number is 12.79∠54.1°. • understand the polar form []r,θ of a complex number and its algebra; ... 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