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RahulZ

  • 4 years ago

find the modulus of the complex number 2+i/3-i

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  1. priyab2
    • 4 years ago
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    first make the complex number with no i in the denominator

  2. priyab2
    • 4 years ago
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    \[(2+i \div3-i) \times (3+i \div3+i)\]

  3. priyab2
    • 4 years ago
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    ie. times by its conjugate

  4. priyab2
    • 4 years ago
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    you get half+half(i)

  5. priyab2
    • 4 years ago
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    then \[\sqrt{0.5+0.5^{2}}\]

  6. priyab2
    • 4 years ago
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    sorry thats wrong \[\sqrt{0.5^{2}+0.5^{2}}\]

  7. priyab2
    • 4 years ago
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    and thats ur mod

  8. adhirajmajumder
    • 4 years ago
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    (2+i)/(3-i) =(2+i)(3+i)/(3+i)(3-i) =(6+5i-1)/9+1 =(5+5i)/10 =(1+i)/2 IZI=\[\sqrt{(1/2)^2+(1/2)^2}\] =\[\sqrt{.25+.25}\] =\[\sqrt{.50}\] =0.7071 \[\tan \theta=1/2/1/2\] \[\tan \theta=\Pi/4\]

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