{"id":851,"date":"2020-09-21T21:33:28","date_gmt":"2020-09-21T21:33:28","guid":{"rendered":"http:\/\/adrianbell.me\/?p=851"},"modified":"2020-09-30T11:02:18","modified_gmt":"2020-09-30T11:02:18","slug":"complex-functions-domains-image-sets-and-inverses","status":"publish","type":"post","link":"https:\/\/adrianbell.me\/?p=851","title":{"rendered":"Complex Functions: Domains, Image Sets and Inverses"},"content":{"rendered":"

<\/p>\n

I can imagine having to refer to these notes regularly, so I'm putting them here!<\/p>\n

Image Sets<\/h1>\n
    \n
  1. State the domain \"A\" of \"f(z)=w\".<\/li>\n
  2. Rearrange so \"w\" is a function of \"z\" (to discover the condition under which \"w\" remains valid.<\/li>\n<\/ol>\n

    e.g., for \"f(z)=\frac{1}{z-1}\":<\/p>\n

      <\/span>   <\/span>\"\begin{align*}<\/p><\/p>\n\n\n

    \n\n\n\n

    Domain of Combined Functions<\/h1>\n\n\n\n

    Domain of combined functions are the intersection (\"A\cap B\") of the domains of all component functions and that of the combined function. e.g:<\/p>\n\n\n\n

    \"f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{Z}-\{0\}\"<\/p>\n\n\n\n

    \"g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{Z}-\{1\}\"<\/p>\n\n\n\n

    \"\frac{f\left(z\right)}{g\left(z\right)} = \frac{z^{2}-2z+1}{z^{2}} \:,\:\: z\in\mathbb{Z}-\{0,\:1\}\"<\/p>\n\n\n\n

    \n\n\n\n

    Domain of Composite Functions<\/h1>\n\n\n\n

    For \"f\" and \"g\" with domains \"A\" and \"B\" respectively, the domain of \"g\circ f\" is:<\/p>\n\n\n\n

    \"A-\{z\::\: f(z)\: \notin\: B\}\"<\/p>\n\n\n\n

    e.g., for:<\/p>\n\n\n\n

    \"f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{C}-\{0\}\"<\/p>\n\n\n\n

    \"g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{C}-\{1\}\"<\/p>\n\n\n\n

    \"\text{domain of }f\circ g = \text{domain of }g - \{z\::\:\frac{z}{z-1} \:\notin\: \mathbb{C}-\{0\}\}\"<\/p>\n\n\n\n

    \"\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{z\::\:\frac{z}{z-1} =0\}\"<\/p>\n\n\n\n

    \"\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{0\}\"<\/p>\n\n\n\n

    \"\text{domain of }f\circ g = (\mathbb{C}-\{0,\:1\} )\"<\/p>\n\n\n\n

    \n\n\n\n

    Inverses<\/h1>\n\n\n\n
    1. Determine image set of \"f(z)=w\".<\/li>
    2. Invert \"f(z)\" to find a unique \"z\" in the domain of \"f\".<\/li><\/ol>\n\n\n\n

      For \"f(z)=\frac{1}{z-1}\"<\/p>\n\n\n\n

      \"f(A) = \{\frac{1}{z-1}\::\:z\in\mathbb{C}-\{1\}\}\"<\/p>\n\n\n\n

      \"f(A) = \{w=\frac{1}{z-1}\::\:z\:\neq\: 1\}\"<\/p>\n\n\n\n

      \"f(A) = \{w\::\: z=\frac{1}{w}+1\:\neq\: 1\}\"<\/p>\n\n\n\n

      \"f(A) = \{w\::\: w\:\neq\: 0\}\"<\/p>\n\n\n\n

      \"f(A) = \mathbb{C}-\{0\}\"<\/p>\n\n\n\n

      (all same as above for finding an image set)<\/p>\n\n\n\n

      \"z=\frac{1}{w}+1\" gives a unique soluition in \"\mathbb{C}-\{0\}\", hence \"f\" has a unique inverse rule:<\/p>\n\n\n\n

      \"f^{-1}(z)=\frac{1}{w}+1\:,\:\:\:z\in\mathbb{C}-\{0\}\"<\/p>\n","protected":false},"excerpt":{"rendered":"

      I can imagine having to refer to these notes regularly, so I'm putting them here! Image Sets State the domain of . Rearrange so is a function of (to discover the condition under which remains valid. e.g., for :     Domain of Combined Functions Domain of combined functions are the intersection () of the … Continue reading Complex Functions: Domains, Image Sets and Inverses<\/span> →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[18,46,14,32],"tags":[47,22,45],"class_list":["post-851","post","type-post","status-publish","format-standard","hentry","category-analysis","category-complex-analysis","category-proofs","category-pure-maths","tag-complex-analysis","tag-proofs","tag-pure-maths"],"_links":{"self":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/851","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=851"}],"version-history":[{"count":26,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/851\/revisions"}],"predecessor-version":[{"id":880,"href":"https:\/\/adrianbell.me\/index.php?rest_route=\/wp\/v2\/posts\/851\/revisions\/880"}],"wp:attachment":[{"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=851"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=851"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/adrianbell.me\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=851"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}