{"id":6525,"date":"2021-10-18T00:00:36","date_gmt":"2021-10-17T18:30:36","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6525"},"modified":"2021-11-21T18:03:42","modified_gmt":"2021-11-21T12:33:42","slug":"polar-form-of-complex-numbers","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/polar-form-of-complex-numbers\/","title":{"rendered":"Polar Form of Complex Numbers with Examples"},"content":{"rendered":"
Here you will learn what is the polar form of complex numbers with examples.<\/p>\n
Let’s begin –\u00a0<\/p>\n
Let z = x + iy be a complex number represented by a point P(x, y) in the argand plane.<\/p>\n
Then by the geometrical representation of z = x + iy, we have<\/p>\n
OP = | z |\u00a0 and angle POX = \\(\\theta\\) = arg (z)<\/p>\n
In triangle POM, we have<\/p>\n
\\(cos\\theta\\) = \\(OM\\over OP\\) = \\(x\\over | z |\\) \\(\\implies\\) x = | z | \\(cos\\theta\\)<\/p>\n
and, \\(sin\\theta\\) = \\(PM\\over OP\\) = \\(y\\over | z |\\) \\(\\implies\\) x = | z | \\(sin\\theta\\)<\/p>\n
So,\u00a0 z = x + iy<\/p>\n
z = | z | \\(cos\\theta\\) + i | z | \\(sin\\theta\\)<\/p>\n
\\(\\implies\\)\u00a0 z = | z | \\((cos\\theta + i sin\\theta)\\)<\/p>\n
\n\\(\\implies\\) z = r\\((cos\\theta + i sin\\theta)\\), where r = | z | and \\(\\theta\\) = arg (z)<\/p>\n<\/blockquote>\n
This form of z is called a polar form of z. If we use the general value of the argument of \\(\\theta\\), then the polar form of z is given by<\/p>\n
\nz = r\\((cos(2n\\pi + \\theta) + i sin(2n\\pi + \\theta))\\), where r = | z | and \\(\\theta\\) = arg (z) and n is an integer.<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Write the following complex numbers in the polar form :\u00a0<\/p>\n
(i) \\(-3\\sqrt{2} + 3\\sqrt{2}i\\)<\/p>\n
(ii) 1 + i<\/p>\n
Solution<\/span><\/strong> :<\/p>\n
(i) Let z = \\(-3\\sqrt{2} + 3\\sqrt{2}i\\). Then,<\/p>\n
r = | z | = \\(\\sqrt{(-3\\sqrt{2})^2 + (3\\sqrt{2})^2}\\) = 6<\/p>\n
\u00a0Let \\(tan\\alpha\\) = |\\(Im(z)\\over Re(z)\\)| = 1 \\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 4\\)<\/p>\n
Since the point representing z lies in the second quadrant. Therefore, the argument of z is given by<\/p>\n
\\(\\theta\\) = \\(\\pi – \\alpha\\) = \\(\\pi\\) – \\(\\pi\\over 4\\) = \\(3\\pi\\over 4\\).<\/p>\n
So, the polar form of z is<\/p>\n
z = r\\((cos\\theta + i sin\\theta)\\) = 6\\(cos{3\\pi\\over 4} + isin{3\\pi\\over 4}\\)<\/p>\n
(ii) Let z = 1 + i. Then,<\/p>\n
r = | z | = \\(\\sqrt{1^2 + 1^2}\\) = \\(\\sqrt{2}\\)<\/p>\n
\u00a0Let \\(tan\\alpha\\) = |\\(Im(z)\\over Re(z)\\)| = 1 \\(\\implies\\) \\(\\alpha\\) = \\(\\pi\\over 4\\)<\/p>\n
Since the point (1, 1) representing z lies in the first quadrant. Therefore, the argument of z is given by<\/p>\n
\\(\\theta\\) = \\(\\alpha\\) = \\(\\pi\\over 4\\)\u00a0<\/p>\n
So, the polar form of z is<\/p>\n
z = r\\((cos\\theta + i sin\\theta)\\) = \\(\\sqrt{2}\\)\\(cos{\\pi\\over 4} + isin{\\pi\\over 4}\\)<\/p>\n\n\n