Products and Quotients of Complex Numbers. The video shows how to divide complex numbers in cartesian form. Hence $\theta = -\dfrac{\pi}{3}+2\pi=\dfrac{5\pi}{3}$ which meets the condition $\theta = \tan^{-1}{\left(\sqrt{3}\right)}$ and also is in the fourth quadrant. We also share information about your use of our site with our social media, advertising and analytics partners. Division of Complex Numbers \[\LARGE \frac{(a+bi)}{(c+di)}=\frac{a+bi}{c+di}\times\frac{c-di}{c-di}=\frac{ac+bd}{c^{2}+d^{2}}+\frac{bc-ad}{c^{2}+d^{2}}i\] Powers of Complex Numbers ), (In this statement, θ is expressed in radian), (We multiplied denominator and numerator with the conjugate of the denominator to proceed), (∵The complex number is in second quadrant), $w_k$ $=r^{1/n}\left[\cos\left(\dfrac{\theta + 2\pi k }{n}\right)+i\sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right]$, (If θ is in degrees, substitute 360° for $2\pi$), $w_k\\=r^{1/n}\left[\cos\left(\dfrac{\theta + 2\pi k}{n}\right)+i\sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right]\\=32^{1/5}\left[\cos\left(\dfrac{\dfrac{\pi}{2}+2\pi k }{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+2\pi k}{5}\right)\right]\\=2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+2\pi k }{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+2\pi k}{5}\right)\right]$, $w_k\\=r^{1/n}\left[\cos\left(\dfrac{\theta + 360°k }{n}\right)+i\sin\left(\dfrac{\theta + 360°k}{n}\right)\right]\\=8^{1/3}\left[\cos\left(\dfrac{\text{240°+360°k}}{3}\right)+i\sin\left(\dfrac{\text{240°+360°k}}{3}\right)\right]\\=2\left[\cos\left(\dfrac{240°+ 360°k }{3}\right)+i\sin\left(\dfrac{240° + 360°k}{3}\right)\right]$, $w_k\\=r^{1/n}\left[\cos\left(\dfrac{\theta + 360°k }{n}\right)+i\sin\left(\dfrac{\theta + 360°k}{n}\right)\right]\\\\=1^{1/3}\left[\cos\left(\dfrac{\text{0°+360°k}}{3}\right)+i\sin\left(\dfrac{\text{0°+360°k}}{3}\right)\right]\\=\cos (120°k)+i\sin (120°k)$. if $z=a+ib$ is a complex number, a is called the real part of z and b is called the imaginary part of z. Conjugate of the complex number $z=x+iy$ can be defined as $\bar{z} = x - iy$, if the complex number $a + ib = 0$, then $a = b = 0$, if the complex number $a + ib = x + iy$, then $a = x$ and $b = y$, if $x + iy$ is a complex numer, then the non-negative real number $\sqrt{x^2 + y^2}$ is the modulus (or absolute value or magnitude) of the complex number $x + iy$. Complex numbers are often denoted by z. (This is because we just add real parts then add imaginary parts; or subtract real parts, subtract imaginary parts.) Maths Formulas - Class XII | Class XI | Class X | Class IX | Class VIII | Class VII | Class VI | Class V Algebra | Set Theory | Trigonometry | Geometry | Vectors | Statistics | Mensurations | Probability | Calculus | Integration | Differentiation | Derivatives Hindi Grammar - Sangya | vachan | karak | Sandhi | kriya visheshan | Vachya | Varnmala | Upsarg | Vakya | Kaal | Samas | kriya | Sarvanam | Ling. Algebraic Structure of Complex Numbers; Division of Complex Numbers; Useful Identities Among Complex Numbers; Useful Inequalities Among Complex Numbers; Trigonometric Form of Complex Numbers So I want to get some real number plus some imaginary number, so some multiple of i's. Addition, subtraction, multiplication and division can be carried out on complex numbers in either rectangular form or polar form. Division of Complex Numbers in Polar Form, Example: Find $\dfrac{5\angle 135° }{4\angle 75°}$, $\dfrac{5\angle 135° }{4\angle 75°} =\dfrac{5}{4}\angle\left( 135° - 75°\right) =\dfrac{5}{4}\angle 60° $, $r=\sqrt{\left(-1\right)^2 +\left(\sqrt{3}\right)^2}\\=\sqrt{1 + 3}=\sqrt{4} = 2$, $\theta = \tan^{-1}{\left(\dfrac{\sqrt{3}}{-1}\right)} = \tan^{-1}{\left(-\sqrt{3}\right)}\\=\dfrac{2\pi}{3}$ (∵The complex number is in second quadrant), $\left(2 \angle 135°\right)^5 = 2^5\left(\angle 135° \times 5\right)\\= 32 \angle 675° = 32 \angle -45°\\=32\left[\cos (-45°)+i\sin (-45°)\right]\\=32\left[\cos (45°) - i\sin (45°)\right]\\= 32\left(\dfrac{1}{\sqrt{2}}-i \dfrac{1}{\sqrt{2}}\right)\\=\dfrac{32}{\sqrt{2}}(1-i)$, $\left[4\left(\cos 30°+i\sin 30°\right)\right]^6 \\= 4^6\left[\cos\left(30° \times 6\right)+i\sin\left(30° \times 6\right)\right]\\=4096\left(\cos 180°+i\sin 180°\right)\\=4096(-1+i\times 0)\\=4096 \times (-1)\\=-4096$, $\left(2e^{0.3i}\right)^8 = 2^8e^{\left(0.3i \times 8\right)} = 256e^{2.4i}\\=256(\cos 2.4+i\sin 2.4)$, $32i = 32\left(\cos \dfrac{\pi}{2}+i\sin \dfrac{\pi}{2}\right)\quad$ (converted to polar form, reference), The 5th roots of 32i can be given by$w_k\\=r^{1/n}\left[\cos\left(\dfrac{\theta + 2\pi k}{n}\right)+i\sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right]\\=32^{1/5}\left[\cos\left(\dfrac{\dfrac{\pi}{2}+2\pi k }{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+2\pi k}{5}\right)\right]\\=2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+2\pi k }{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+2\pi k}{5}\right)\right]$, $w_0 = 2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+0}{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+0}{5}\right)\right]$ $= 2\left(\cos \dfrac{\pi}{10}+i\sin \dfrac{\pi}{10}\right)$, $w_1 = 2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+2\pi}{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+2\pi}{5}\right)\right]$ $= 2\left(\cos \dfrac{\pi}{2}+i\sin \dfrac{\pi}{2}\right) = 2i$, $w_2 = 2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+4\pi}{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+4\pi}{5}\right)\right]$ $= 2\left(\cos \dfrac{9\pi}{10}+i\sin \dfrac{9\pi}{10}\right)$, $w_3 = 2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+6\pi}{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+6\pi}{5}\right)\right]$ $= 2\left(\cos \dfrac{13\pi}{10}+i\sin \dfrac{13\pi}{10}\right)$, $w_4 = 2\left[\cos\left(\dfrac{\dfrac{\pi}{2}+8\pi}{5}\right)+i\sin\left(\dfrac{\dfrac{\pi}{2}+8\pi}{5}\right)\right]$ $= 2\left(\cos \dfrac{17\pi}{10}+i\sin \dfrac{17\pi}{10}\right)$, $-4 - 4\sqrt{3}i = 8\left(\cos 240°+i\sin 240°\right)\quad$(converted to polar form, reference. 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