So, \[\dfrac{w}{z} = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha) + i\sin(\alpha))}{(\cos(\beta) + i\sin(\beta)} \right ] = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha) + i\sin(\alpha))}{(\cos(\beta) + i\sin(\beta)} \cdot \dfrac{(\cos(\beta) - i\sin(\beta))}{(\cos(\beta) - i\sin(\beta)} \right ] = \dfrac{r}{s}\left [\dfrac{(\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)) + i(\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)}{\cos^{2}(\beta) + \sin^{2}(\beta)} \right ]\]. Also, \(|z| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}\) and the argument of \(z\) is \(\arctan(\dfrac{-1}{1}) = -\dfrac{\pi}{4}\). 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. But in polar form, the complex numbers are represented as the combination of modulus and argument. a =-2 b =-2. Let and be two complex numbers in polar form. We have seen that we multiply complex numbers in polar form by multiplying their norms and adding their arguments. Therefore, if we add the two given complex numbers, we get; Again, to convert the resulting complex number in polar form, we need to find the modulus and argument of the number. The angle \(\theta\) is called the argument of the complex number \(z\) and the real number \(r\) is the modulus or norm of \(z\). So, \[\dfrac{w}{z} = \dfrac{r(\cos(\alpha) + i\sin(\alpha))}{s(\cos(\beta) + i\sin(\beta)} = \dfrac{r}{s}\left [\dfrac{\cos(\alpha) + i\sin(\alpha)}{\cos(\beta) + i\sin(\beta)} \right ]\], We will work with the fraction \(\dfrac{\cos(\alpha) + i\sin(\alpha)}{\cos(\beta) + i\sin(\beta)}\) and follow the usual practice of multiplying the numerator and denominator by \(\cos(\beta) - i\sin(\beta)\). Back to the division of complex numbers in polar form. How to algebraically calculate exact value of a trig function applied to any non-transcendental angle? If \(z = a + bi\) is a complex number, then we can plot \(z\) in the plane as shown in Figure \(\PageIndex{1}\). There is a similar method to divide one complex number in polar form by another complex number in polar form. Khan Academy is a 501(c)(3) nonprofit organization. Step 1. How do we multiply two complex numbers 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). Products and Quotients of Complex Numbers. Thanks to all of you who support me on Patreon. What is the polar (trigonometric) form of a complex number? Recall that \(\cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2}\) and \(\sin(\dfrac{\pi}{6}) = \dfrac{1}{2}\). 4. Multiplication of complex numbers is more complicated than addition of complex numbers. \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), 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. Let \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) be complex numbers in polar form with \(z \neq 0\). 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 . \[|\dfrac{w}{z}| = \dfrac{|w|}{|z|} = \dfrac{3}{2}\], 2. Note that \(|w| = \sqrt{4^{2} + (4\sqrt{3})^{2}} = 4\sqrt{4} = 8\) and the argument of \(w\) is \(\arctan(\dfrac{4\sqrt{3}}{4}) = \arctan\sqrt{3} = \dfrac{\pi}{3}\). 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. 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. The following development uses trig.formulae you will meet in Topic 43. If a n = b, then a is said to be the n-th root of b. To convert into polar form modulus and argument of the given complex number, i.e. Your email address will not be published. How to solve this? 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. 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 \(wz\). We won’t go into the details, but only consider this as notation. \[^* \space \theta = \dfrac{\pi}{2} \space if \space b > 0\] Complex Number Division Formula, what is a complex number, roots of complex numbers, magnitude of complex number, operations with complex numbers. (This is spoken as “r at angle θ ”.) 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 \[\dfrac{5\pi}{3} - \dfrac{\pi}{4} = \dfrac{20\pi - 3\pi}{12} = \dfrac{17\pi}{12}\]. This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers. Determine the polar form of \(|\dfrac{w}{z}|\). The argument of \(w\) is \(\dfrac{5\pi}{3}\) and the argument of \(z\) is \(-\dfrac{\pi}{4}\), we see that the argument of \(\dfrac{w}{z}\) is, \[\dfrac{5\pi}{3} - (-\dfrac{\pi}{4}) = \dfrac{20\pi + 3\pi}{12} = \dfrac{23\pi}{12}\]. 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. Your email address will not be published. We know the magnitude and argument of \(wz\), so the polar form of \(wz\) is, \[wz = 6[\cos(\dfrac{17\pi}{12}) + \sin(\dfrac{17\pi}{12})]\]. 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 \(z \neq 0\). This turns out to be true in general. Since \(w\) is in the second quadrant, we see that \(\theta = \dfrac{2\pi}{3}\), so the polar form of \(w\) is \[w = \cos(\dfrac{2\pi}{3}) + i\sin(\dfrac{2\pi}{3})\]. 1. With Euler’s formula we can rewrite the polar form of a complex number into its exponential form as follows. Answer: ... How do I find the quotient of two complex numbers in polar form? 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, \[\dfrac{w}{z} = \dfrac{r}{s}(\cos(\alpha - \beta) + i\sin(\alpha - \beta)),\]. We know the magnitude and argument of \(wz\), so the polar form of \(wz\) is \[\dfrac{w}{z} = \dfrac{3}{2}[\cos(\dfrac{23\pi}{12}) + \sin(\dfrac{23\pi}{12})]\], Let \(w = r(\cos(\alpha) + i\sin(\alpha))\) and \(z = s(\cos(\beta) + i\sin(\beta))\) be complex numbers in polar form with \(z \neq 0\). If \(z \neq 0\) and \(a \neq 0\), then \(\tan(\theta) = \dfrac{b}{a}\). We now use the following identities with the last equation: Using these identities with the last equation for \(\dfrac{w}{z}\), we see that, \[\dfrac{w}{z} = \dfrac{r}{s}[\dfrac{\cos(\alpha - \beta) + i\sin(\alpha- \beta)}{1}].\]. Following is a picture of \(w, z\), and \(wz\) that illustrates the action of the complex product. The polar form of a complex number z = a + b i is z = r ( cos θ + i sin θ ) , where r = | z | = a 2 + b 2 , a = r cos θ and b = r sin θ , and θ = tan − 1 ( b a ) for a > 0 and θ = tan − 1 … The complex conjugate of a complex number can be found by replacing the i in equation [1] with -i. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. Usually, we represent the complex numbers, in the form of z = x+iy where ‘i’ the imaginary number.But in polar form, the complex numbers are represented as the combination of modulus and argument. ... A Complex number is in the form of a+ib, where a and b are real numbers the ‘i’ is called the imaginary unit. 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 \(z\). The following applets demonstrate what is going on when we multiply and divide complex numbers. r 2 cis θ 2 = r 1 r 2 (cis θ 1 . 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 \(z\). 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. There is an important product formula for complex numbers that the polar form provides. Complex numbers are often denoted by z. Let us learn here, in this article, how to derive the polar form of complex numbers. \[e^{i\theta} = \cos(\theta) + i\sin(\theta)\] There is an alternate representation that you will often see for the polar form of a complex number using a complex exponential. $1 per month helps!! You da real mvps! Draw a picture of \(w\), \(z\), and \(|\dfrac{w}{z}|\) that illustrates the action of the complex product. divide them. ⇒ z1z2 = r1eiθ1. \[^* \space \theta = -\dfrac{\pi}{2} \space if \space b < 0\], 1. 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. Multiply & divide complex numbers in polar form Our mission is to provide a free, world-class education to anyone, anywhere. If \(z \neq 0\) and \(a = 0\) (so \(b \neq 0\)), then. 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! This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers. So \(a = \dfrac{3\sqrt{3}}{2}\) and \(b = \dfrac{3}{2}\). Using our definition of the product of complex numbers we see that, \[wz = (\sqrt{3} + i)(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i) = -\sqrt{3} + i.\] Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. 5. Required fields are marked *. What is the argument of \(|\dfrac{w}{z}|\)? 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. 4. Since \(|w| = 3\) and \(|z| = 2\), we see that, 2. 1. 4. Determine the conjugate of the denominator. The following questions are meant to guide our study of the material in this section. Proof that unit complex numbers 1, z and w form an equilateral triangle. Division of complex numbers means doing the mathematical operation of division on complex numbers. Indeed, using the product theorem, (z1 z2)⋅ z2 = {(r1 r2)[cos(ϕ1 −ϕ2)+ i⋅ sin(ϕ1 −ϕ2)]} ⋅ r2(cosϕ2 +i ⋅ sinϕ2) = Using equation (1) and these identities, we see that, \[w = rs([\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)]) + i[\cos(\alpha)\sin(\beta) + \cos(\beta)\sin(\alpha)] = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\]. Example If z What is the complex conjugate of a complex number? 3. This states that to multiply two complex numbers in polar form, we multiply their norms and add their arguments. The rectangular form of a complex number is denoted by: In the case of a complex number, r signifies the absolute value or modulus and the angle θ is known as the argument of the complex number. For longhand multiplication and division, polar is the favored notation to work with. [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. (This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts.) Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers. In which quadrant is \(|\dfrac{w}{z}|\)? The proof of this is similar to the proof for multiplying complex numbers and is included as a supplement to this section. z = r z e i θ z. z = r_z e^{i \theta_z}. If \(r\) is the magnitude of \(z\) (that is, the distance from \(z\) to the origin) and \(\theta\) the angle \(z\) makes with the positive real axis, then the trigonometric form (or polar form) of \(z\) is \(z = r(\cos(\theta) + i\sin(\theta))\), where, \[r = \sqrt{a^{2} + b^{2}}, \cos(\theta) = \dfrac{a}{r}\]. Then the polar form of the complex product \(wz\) is given by, \[wz = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))\]. rieiθ2 = r1r2ei(θ1+θ2) ⇒ z 1 z 2 = r 1 e i θ 1. r i e i θ 2 = r 1 r 2 e i ( θ 1 + θ 2) This result is in agreement with the fact that moduli multiply and arguments add upon multiplication. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . The parameters \(r\) and \(\theta\) are the parameters of the polar form. Missed the LibreFest? To find \(\theta\), we have to consider cases. Exercise \(\PageIndex{13}\) The result of Example \(\PageIndex{1}\) is no coincidence, as we will show. 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