{"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
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.<\/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 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\nAlgorithm :<\/strong><\/h4>\n
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