This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers. The proof of this is similar to the proof for multiplying complex numbers and is included as a supplement to this section. After studying this section, we should understand the concepts motivated by these questions and be able to write precise, coherent answers to these questions. What is the argument of $$|\dfrac{w}{z}|$$? Hence. Let z 1 = r 1 cis θ 1 and z 2 = r 2 cis θ 2 be any two complex numbers. An illustration of this is given in Figure $$\PageIndex{2}$$. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Multiplication of complex numbers is more complicated than addition of complex numbers. Required fields are marked *. Draw a picture of $$w$$, $$z$$, and $$|\dfrac{w}{z}|$$ that illustrates the action of the complex product. Now we write $$w$$ and $$z$$ in polar form. 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 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. Back to the division of complex numbers in polar form. N-th root of a number. 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. So $$a = \dfrac{3\sqrt{3}}{2}$$ and $$b = \dfrac{3}{2}$$. The modulus of a complex number is also called absolute value. Exercise $$\PageIndex{13}$$ This turns out to be true in general. 1. Your email address will not be published. This is an advantage of using the polar form. Since $$|w| = 3$$ and $$|z| = 2$$, we see that, 2. This polar form is represented with the help of polar coordinates of real and imaginary numbers in the coordinate system. 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 $$\cos(\beta) - i\sin(\beta)$$. Based on this definition, complex numbers can be added and … The following figure shows the complex number z = 2 + 4j Polar and exponential form. 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. Complex numbers are built on the concept of being able to define the square root of negative one. Note that $$|w| = \sqrt{(-\dfrac{1}{2})^{2} + (\dfrac{\sqrt{3}}{2})^{2}} = 1$$ and the argument of $$w$$ satisfies $$\tan(\theta) = -\sqrt{3}$$. 4. But in polar form, the complex numbers are represented as the combination of modulus and argument. Multiply the numerator and denominator by the conjugate . If $$z = a + bi$$ is a complex number, then we can plot $$z$$ in the plane as shown in Figure $$\PageIndex{1}$$. $^* \space \theta = \dfrac{\pi}{2} \space if \space b > 0$ Roots of complex numbers in polar form. This video gives the formula for multiplication and division of two complex numbers that are in polar form… z =-2 - 2i z = a + bi, Let us consider (x, y) are the coordinates of complex numbers x+iy. Multiplication. Let $$w = r(\cos(\alpha) + i\sin(\alpha))$$ and $$z = s(\cos(\beta) + i\sin(\beta))$$ be complex numbers in polar form with $$z \neq 0$$. Explain. When we write $$e^{i\theta}$$ (where $$i$$ is the complex number with $$i^{2} = -1$$) we mean. Let $$w = 3[\cos(\dfrac{5\pi}{3}) + i\sin(\dfrac{5\pi}{3})]$$ and $$z = 2[\cos(-\dfrac{\pi}{4}) + i\sin(-\dfrac{\pi}{4})]$$. How to algebraically calculate exact value of a trig function applied to any non-transcendental angle? $$\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) = \cos(\alpha - \beta)$$, $$\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) = \sin(\alpha - \beta)$$, $$\cos^{2}(\beta) + \sin^{2}(\beta) = 1$$. The result of Example $$\PageIndex{1}$$ is no coincidence, as we will show. The terminal side of an angle of $$\dfrac{17\pi}{12} = \pi + \dfrac{5\pi}{12}$$ radians is in the third quadrant. If $$z \neq 0$$ and $$a = 0$$ (so $$b \neq 0$$), then. The formula for multiplying complex numbers in polar form tells us that to multiply two complex numbers, we add their arguments and multiply their norms. 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. 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. Have questions or comments? This trigonometric form connects algebra to trigonometry and will be useful for quickly and easily finding powers and roots of complex numbers. The following questions are meant to guide our study of the material in this section. 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, $\dfrac{w}{z} = \dfrac{r}{s}(\cos(\alpha - \beta) + i\sin(\alpha - \beta)),$. Your email address will not be published. To better understand the product of complex numbers, we first investigate the trigonometric (or polar) form of a complex number. z = r z e i θ z. z = r_z e^{i \theta_z}. • understand the polar form []r,θ of a complex number and its algebra; ... Activity 6 Division Simplify to the form a +ib (a) 4 i (b) 1−i 1+i (c) 4 +5i 6 −5i (d) 4i ()1+2i 2 3.2 Solving equations Just as you can have equations with real numbers, you can have Determine real numbers $$a$$ and $$b$$ so that $$a + bi = 3(\cos(\dfrac{\pi}{6}) + i\sin(\dfrac{\pi}{6}))$$. divide them. 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. a =-2 b =-2. Every complex number can also be written in polar form. 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! 1. Convert given two complex number division into polar form. Use right triangle trigonometry to write $$a$$ and $$b$$ in terms of $$r$$ and $$\theta$$. z 1 z 2 = r 1 cis θ 1 . Example $$\PageIndex{1}$$: Products of Complex Numbers in Polar Form, Let $$w = -\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i$$ and $$z = \sqrt{3} + i$$. To better understand the product of complex numbers, we first investigate the trigonometric (or polar) form of a complex number. Let us learn here, in this article, how to derive the polar form of complex numbers. The complex conjugate of a complex number can be found by replacing the i in equation [1] with -i. (This is spoken as “r at angle θ ”.) … Using our definition of the product of complex numbers we see that, $wz = (\sqrt{3} + i)(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i) = -\sqrt{3} + i.$ The terminal side of an angle of $$\dfrac{23\pi}{12} = 2\pi - \dfrac{\pi}{12}$$ radians is in the fourth quadrant. Example: Find the polar form of complex number 7-5i. To convert into polar form modulus and argument of the given complex number, i.e. Let 3+5i, and 7∠50° are the two complex numbers. \$1 per month helps!! Missed the LibreFest? For longhand multiplication and division, polar is the favored notation to work with. Following is a picture of $$w, z$$, and $$wz$$ that illustrates the action of the complex product. The conjugate of ( 7 + 4 i) is ( 7 − 4 i) . 5 + 2 i 7 + 4 i. The graphical representation of the complex number $$a+ib$$ is shown in the graph below. Multiplication and Division of Complex Numbers in Polar Form Complex numbers are often denoted by z. Division of complex numbers means doing the mathematical operation of division on complex numbers. 1. Then the polar form of the complex product $$wz$$ is given by, $wz = rs(\cos(\alpha + \beta) + i\sin(\alpha + \beta))$. 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 . 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, $$|z| = \sqrt{(\sqrt{3})^{2} + 1^{2}} = 2$$ and the argument of $$z$$ satisfies $$\tan(\theta) = \dfrac{1}{\sqrt{3}}$$. Legal. Indeed, using the product theorem, (z1 z2)⋅ z2 = {(r1 r2)[cos(ϕ1 −ϕ2)+ i⋅ sin(ϕ1 −ϕ2)]} ⋅ r2(cosϕ2 +i ⋅ sinϕ2) = Determine the polar form of $$|\dfrac{w}{z}|$$. $^* \space \theta = -\dfrac{\pi}{2} \space if \space b < 0$, 1. How do we multiply two complex numbers in polar form? by M. Bourne. Free Complex Number Calculator for division, multiplication, Addition, and Subtraction So, $w = 8(\cos(\dfrac{\pi}{3}) + \sin(\dfrac{\pi}{3}))$. In this situation, we will let $$r$$ be the magnitude of $$z$$ (that is, the distance from $$z$$ to the origin) and $$\theta$$ the angle $$z$$ makes with the positive real axis as shown in Figure $$\PageIndex{1}$$. If $$z \neq 0$$ and $$a \neq 0$$, then $$\tan(\theta) = \dfrac{b}{a}$$. Step 1. Note that $$|w| = \sqrt{4^{2} + (4\sqrt{3})^{2}} = 4\sqrt{4} = 8$$ and the argument of $$w$$ is $$\arctan(\dfrac{4\sqrt{3}}{4}) = \arctan\sqrt{3} = \dfrac{\pi}{3}$$. 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. 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 $$z \neq 0$$. $|\dfrac{w}{z}| = \dfrac{|w|}{|z|} = \dfrac{3}{2}$, 2. 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 $$z$$. by M. Bourne. Example If z 5. 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 $$wz$$. r and θ. How do we divide one complex number in polar form by a nonzero complex number in polar form? Answer: ... How do I find the quotient of two complex numbers in polar form? There is a similar method to divide one complex number in polar form by another complex number in polar form. Solution:7-5i is the rectangular form of a complex number. Recall that $$\cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2}$$ and $$\sin(\dfrac{\pi}{6}) = \dfrac{1}{2}$$. To find $$\theta$$, we have to consider cases. If $$z = 0 = 0 + 0i$$,then $$r = 0$$ and $$\theta$$ can have any real value. \]. 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 $$z \neq 0$$. So to divide complex numbers in polar form, we divide the norm of the complex number in the numerator by the norm of the complex number in the denominator and subtract the argument of the complex number in the denominator from the argument of the complex number in the numerator. :) https://www.patreon.com/patrickjmt !! 4. Key Questions. Let and be two complex numbers in polar form. 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. Proof of the Rule for Dividing Complex Numbers in Polar Form. z = r z e i θ z . Polar Form of a Complex Number. So $3(\cos(\dfrac{\pi}{6} + i\sin(\dfrac{\pi}{6})) = 3(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i) = \dfrac{3\sqrt{3}}{2} + \dfrac{3}{2}i$. The following development uses trig.formulae you will meet in Topic 43. We now use the following identities with the last equation: Using these identities with the last equation for $$\dfrac{w}{z}$$, we see that, $\dfrac{w}{z} = \dfrac{r}{s}[\dfrac{\cos(\alpha - \beta) + i\sin(\alpha- \beta)}{1}].$. 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. ⇒ z1z2 = r1eiθ1. 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. Figure $$\PageIndex{2}$$: A Geometric Interpretation of Multiplication of Complex Numbers. To find the polar representation of a complex number $$z = a + bi$$, we first notice that. We have seen that we multiply complex numbers in polar form by multiplying their norms and adding their arguments. 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. Multiplication and division of complex numbers in polar form. Draw a picture of $$w$$, $$z$$, and $$wz$$ that illustrates the action of the complex product. To divide,we divide their moduli and subtract their arguments. $z = r{{\bf{e}}^{i\,\theta }}$ 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. Products and Quotients of Complex Numbers. This states that to multiply two complex numbers in polar form, we multiply their norms and add their arguments. [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. The polar form of a complex number is a different way to represent a complex number apart from rectangular form. 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… 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 $$z$$. If a n = b, then a is said to be the n-th root of b. What is the complex conjugate of a complex number? We illustrate with an example. (This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts.) 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: Then, the product and quotient of these are given by When performing addition and subtraction of complex numbers, use rectangular form. The word polar here comes from the fact that this process can be viewed as occurring with polar coordinates. \[z = r(\cos(\theta) + i\sin(\theta)). The parameters $$r$$ and $$\theta$$ are the parameters of the polar form. In which quadrant is $$|\dfrac{w}{z}|$$? 4. Solution The complex number is in rectangular form with and We plot the number by moving two units to the left on the real axis and two units down parallel to the imaginary axis, as shown in Figure 6.43 on the next page. This is the polar 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. Usually, we represent the complex numbers, in the form of z = x+iy where ‘i’ the imaginary number. 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