{"id":6524,"date":"2021-10-17T23:26:43","date_gmt":"2021-10-17T17:56:43","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6524"},"modified":"2021-11-21T18:07:04","modified_gmt":"2021-11-21T12:37:04","slug":"amplitude-of-a-complex-number","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/amplitude-of-a-complex-number\/","title":{"rendered":"Amplitude of a Complex Number"},"content":{"rendered":"

Here you will learn how to find argument or amplitude of a complex number with examples.<\/p>\n

Let’s begin –\u00a0<\/p>\n

Amplitude of a Complex Number (Argument of Complex Number)<\/h2>\n

Let z = x + iy, Then,<\/p>\n

The angle \\(\\theta\\) which OP makes with the positive direction of x-axis in anticlockwise sense is called the argument or amplitude of complex number z.\"amplitude\"<\/p>\n

It is denoted by arg(z) or amp(z).<\/p>\n

From Figure, we have<\/p>\n

\\(tan\\theta\\) = \\(PM\\over OM\\) = \\(y\\over x\\) = \\(Im (z)\\over Re (z)\\)<\/p>\n

\\(\\implies\\) \\(\\theta\\) = \\(tan^{-1}({y\\over x}\\))<\/p>\n

This angle \\(\\theta\\) has infinitely many values differing by multiples of \\(2\\pi\\). The unique value of \\(\\theta\\) such that -\\(\\pi\\) < \\(\\theta\\) \\(\\le\\) \\(\\pi\\) is called the principal value of the amplitude or principal argument.<\/strong><\/p>\n

In order to find the argument of a complex number z = x + iy for different quadrants, we use the following algorithm.<\/p>\n

Algorithm :<\/strong><\/h4>\n
\n

1). Find the value of \\(tan^{-1}| y\/x |\\) lying between 0 and \\(\\pi\/2\\). Let it be \\(\\alpha\\).<\/p>\n

2). Determine in which quadrant the point P(x, y) belongs.<\/p>\n

(i) If P(x, y ) belongs to the first quadrant, then arg(z) = \\(\\alpha\\).<\/p>\n

(ii) If P(x, y ) belongs to the second quadrant, then arg(z) = \\(\\pi – \\alpha\\).<\/p>\n

(iii) If P(x, y ) belongs to the third quadrant, then arg(z) = -(\\(\\pi – \\alpha\\)) or \\(\\pi + \\alpha\\).<\/p>\n

(iv) If P(x, y ) belongs to the fourth quadrant, then arg(z) = -\\(\\alpha\\) or \\(2\\pi – \\alpha\\).<\/p>\n<\/blockquote>\n

Example<\/strong><\/span> : Find the argument of the following complex numbers.<\/p>\n

(i) \\(1 + i\\sqrt{3}\\)<\/p>\n

(ii) \\(-2 + 2i\\sqrt{3}\\)<\/p>\n

(iii) \\(-\\sqrt{3} – i\\)<\/p>\n

(iv) \\(2\\sqrt{3} – 2i\\)<\/p>\n

Solution<\/span><\/strong> :\u00a0<\/p>\n

(i) Let\u00a0 z = \\(1 + i\\sqrt{3}\\) and let \\(\\alpha\\) be the acute angle given by<\/p>\n

\\(\\alpha\\) = \\(tan^{-1}|{Im(z)\\over Re(z)}|\\) = \\(tan^{-1}|{\\sqrt{3}\\over 1}|\\)<\/p>\n

\\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 3\\)<\/p>\n

Since, Re(z) > 0 and Im(z) > 0. So, it lies in first quadrant\u00a0<\/p>\n

\\(\\therefore\\)\u00a0 arg(z) = \\(\\alpha\\) = \\(\\pi\\over 3\\)<\/p>\n

(ii) Let\u00a0 z = \\(-2 + 2i\\sqrt{3}\\) and let \\(\\alpha\\) be the acute angle given by<\/p>\n

\\(\\alpha\\) = \\(tan^{-1}|{Im(z)\\over Re(z)}|\\) = \\(tan^{-1}|{2\\sqrt{3}\\over -2}|\\)<\/p>\n

\\(\\implies\\) arg(z) = \\(\\alpha\\) = \\(\\pi\\over 3\\)<\/p>\n

Since, Re(z) < 0 and Im(z) > 0. So, it lies in second quadrant\u00a0<\/p>\n

\\(\\therefore\\) arg(z) = \\(\\pi – \\alpha\\) = \\(\\pi\\) – \\(\\pi\\over 3\\) = \\(2\\pi\\over 3\\)<\/p>\n

(iii) Let\u00a0 z = \\(-\\sqrt{3} – i\\) and let \\(\\alpha\\) be the acute angle given by<\/p>\n

\\(\\alpha\\) = \\(tan^{-1}|{Im(z)\\over Re(z)}|\\) = \\(tan^{-1}|{-1\\over -\\sqrt{3}}|\\)<\/p>\n

\\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 6\\)<\/p>\n

Since, Re(z) < 0 and Im(z) < 0. So, it lies in third quadrant\u00a0<\/p>\n

\\(\\therefore\\)\u00a0 arg(z) = -(\\(\\pi – \\alpha\\)) = -(\\(\\pi\\) – \\(\\pi\\over 6\\)) = \\(-5\\pi\\over 6\\)<\/p>\n

(iv) Let\u00a0 z = \\(2\\sqrt{3} – 2i\\) and let \\(\\alpha\\) be the acute angle given by<\/p>\n

\\(\\alpha\\) = \\(tan^{-1}|{Im(z)\\over Re(z)}|\\) = \\(tan^{-1}|{-2\\over 2\\sqrt{3}}|\\)<\/p>\n

\\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 6\\)<\/p>\n

Since, Re(z) > 0 and Im(z) < 0. So, it lies in fourth quadrant\u00a0<\/p>\n

\\(\\therefore\\)\u00a0 arg(z) = -\\(\\alpha\\) = -\\(\\pi\\over 6\\)<\/p>\n\n\n

\n
Next – Polar Form of Complex Numbers with Examples<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – How to Find Square Root of Complex Number<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn how to find argument or amplitude of a complex number with examples. Let’s begin –\u00a0 Amplitude of a Complex Number (Argument of Complex Number) Let z = x + iy, Then, The angle \\(\\theta\\) which OP makes with the positive direction of x-axis in anticlockwise sense is called the argument or …<\/p>\n

Amplitude of a Complex Number<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[41],"tags":[231,232,229],"yoast_head":"\nAmplitude of a Complex Number - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn how to find argument or amplitude of a complex number with examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/amplitude-of-a-complex-number\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Amplitude of a Complex Number - 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