{"id":6505,"date":"2021-10-17T15:10:45","date_gmt":"2021-10-17T09:40:45","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6505"},"modified":"2021-11-21T18:29:36","modified_gmt":"2021-11-21T12:59:36","slug":"integral-powers-of-iota-complex-numbers","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integral-powers-of-iota-complex-numbers\/","title":{"rendered":"Integral Powers of Iota – Complex Numbers"},"content":{"rendered":"

Here you will learn complex number i (iota) and integral powers of iota with examples.<\/p>\n

Let’s begin –<\/p>\n

Integral Powers of Iota (i)<\/h2>\n

Positive Integral Powers of iota(i) :<\/strong><\/p>\n

We have,<\/p>\n

\n

i = \\(\\sqrt{-1}\\)<\/p>\n

\\(\\therefore\\) \\(i^2\\) = -1<\/p>\n

\\(i^3\\) = \\(i^2\\) \\(\\times\\) i = -i<\/p>\n

\\(i^4\\) = \\((i^2)^2\\) = \\((-1)^2\\) = 1<\/p>\n<\/blockquote>\n

In order to compute \\(i^n\\) for n > 4, we divide n by 4 and obtain the remainder r. Let m be the quotient when n is divided by 4. Then,<\/p>\n

\n

n = 4m + r, where 0 \\(\\le\\) r < 4<\/p>\n

\\(\\implies\\) \\(i^n\\) = \\(i^{4m + r}\\) = \\((i^4)^m i^r\\) = \\(i^4\\)<\/p>\n<\/blockquote>\n

Thus, the value of \\(i^n\\) for n > 4 is \\(i^r\\), where r is the remainder when n is divide by 4.<\/p>\n

Negative integral powers of iota(i) :<\/strong><\/p>\n

By the law of indices, we have<\/p>\n

\n

\\(i^{-1}\\) = \\(1\\over i\\) = \\(i^3\\over i^4\\) = \\(i^3\\) = -i<\/p>\n

\\(i^{-2}\\) = \\(1\\over i^2\\) = \\(1\\over -1\\) = -1<\/p>\n

\\(i^{-3}\\) = \\(1\\over i^3\\) = \\(1\\over i^4\\) = i<\/p>\n

\\(i^{-4}\\) = \\(1\\over i^4\\) = \\(1\\over 1\\) = 1<\/p>\n<\/blockquote>\n

If n > 4, then<\/p>\n

\n

\\(i^{-n}\\) = \\(1\\over i^n\\) = \\(1\\over i^r\\), where r is the remainder when n is divided by 4.<\/p>\n<\/blockquote>\n

Note<\/strong> : \\(i^{0}\\) is defined as 1.<\/p>\n

Example<\/span><\/strong> : Evaluate the following :<\/p>\n

(i) \\(i^{135}\\)           <\/p>\n

(ii) \\(i^{19}\\)<\/p>\n

(iii) \\(i^{-999}\\)<\/p>\n

Solution<\/span><\/strong> : <\/p>\n

(i) 135 leaves remainder as 3 when it is divided by 4.<\/p>\n

\\(\\therefore\\)  \\(i^{135}\\) = \\(i^3\\) = -i<\/p>\n

(ii) The remainder is when 19 is divided by 4.<\/p>\n

\\(\\therefore\\)  \\(i^{19}\\) = \\(i^3\\) = -i<\/p>\n

(iii) We have, \\(i^{-999}\\) = \\(1\\over i^{999}\\)<\/p>\n

On dividing 999 by 4, we obtain 3 as the remainder.<\/p>\n

\\(\\therefore\\) \\(i^{999}\\) = \\(i^3\\)<\/p>\n

\\(\\implies\\) \\(i^{-999}\\) = \\(1\\over i^{999}\\) = \\(1\\over i^3\\) = \\(i\\over i^4\\) = \\(i\\over 1\\) = i<\/p>\n\n\n

\n
Next – Complex Number Class 11<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn complex number i (iota) and integral powers of iota with examples. Let’s begin – Integral Powers of Iota (i) Positive Integral Powers of iota(i) : We have, i = \\(\\sqrt{-1}\\) \\(\\therefore\\) \\(i^2\\) = -1 \\(i^3\\) = \\(i^2\\) \\(\\times\\) i = -i \\(i^4\\) = \\((i^2)^2\\) = \\((-1)^2\\) = 1 In order to …<\/p>\n

Integral Powers of Iota – Complex Numbers<\/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":[247,229,248],"yoast_head":"\nIntegral Powers of Iota - Complex Numbers<\/title>\n<meta name=\"description\" content=\"In this post you will learn complex number i (iota) and integral powers of iota 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\/integral-powers-of-iota-complex-numbers\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Integral Powers of Iota - 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