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JenniferSmart1

why is this true? \[\frac{2}{8i}=-\frac i 4\]

  • one year ago
  • one year ago

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  1. satellite73
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    first off you can reduce, just like with real numbers then multiply top and bottom by \(i\)

    • one year ago
  2. matricked
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    because i*i=-1

    • one year ago
  3. matricked
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    or 1/i=-i

    • one year ago
  4. JenniferSmart1
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    Is our goal to get the imaginary number out of the denominator?

    • one year ago
  5. sauravshakya
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    |dw:1358398893659:dw|

    • one year ago
  6. satellite73
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    \[\frac{2}{8i}=\frac{1}{4i}=\frac{1}{4i}\times \frac{i}{i}=\frac{i}{4i^2}=-\frac{i}{4}\]

    • one year ago
  7. sauravshakya
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    What is wrong if I say 1/i=i

    • one year ago
  8. satellite73
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    it is \(\frac{1}{i}=-i\)

    • one year ago
  9. satellite73
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    @sauravshakya you cannot use \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) unless \(a\) and \(b\) are real

    • one year ago
  10. sauravshakya
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    ya I got how u got 1/i=-i But |dw:1358399037098:dw| this doesn't work

    • one year ago
  11. satellite73
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    no it does not

    • one year ago
  12. sauravshakya
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    1 and -1 are real

    • one year ago
  13. satellite73
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    even this doesn't work \[\sqrt{a}\sqrt{b}=\sqrt{ab}\] unless \(a\) and \(b\) are positive numbers

    • one year ago
  14. satellite73
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    excuse me i said 'real' and i meant 'positive' my mistake

    • one year ago
  15. sauravshakya
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    oh I got it...

    • one year ago
  16. satellite73
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    @JenniferSmart1 yes the goal is to get the complex number out of the denominator to write in standard form \(a+bi\)

    • one year ago
  17. JenniferSmart1
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    oh ok

    • one year ago
  18. JenniferSmart1
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    make sense

    • one year ago
  19. JenniferSmart1
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    so i*i=-1?

    • one year ago
  20. JenniferSmart1
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    I'm too tired...I'll look at it again tomorrow. Thanks guys

    • one year ago
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