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sedighn Group Title

Find the square roots of 5+2i, giving your answers in the form a+bi

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    do you know how to convert the form a+bi into polar form, r <theta ?

    • one year ago
  2. hartnn Group Title
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    and vice-versa ?

    • one year ago
  3. dumbcow Group Title
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    \[\sqrt{5+2i}\] ?

    • one year ago
  4. sedighn Group Title
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    Yes I do... I've never seen such a question so I'm not sure how to begin!

    • one year ago
  5. hartnn Group Title
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    convert 5+2i into polar first

    • one year ago
  6. sedighn Group Title
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    okay so... \[\left[ \sqrt{29},0.381 \right]\]

    • one year ago
  7. sedighn Group Title
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    (in radians)

    • one year ago
  8. dumbcow Group Title
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    \[\sqrt{r(\cos \theta +i \sin \theta)} = \sqrt{r}(\cos \frac{\theta}{2} +i \sin \frac{\theta}{2})\]

    • one year ago
  9. hartnn Group Title
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    yes ^

    • one year ago
  10. hartnn Group Title
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    so just take square root of r half the angle and then convert it back to a+bi form

    • one year ago
  11. sedighn Group Title
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    wow okay thanks a lot! we didn't do this method in class o.o

    • one year ago
  12. hartnn Group Title
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    well, now you are smarter than your classmates ;) (Y)

    • one year ago
  13. jim_thompson5910 Group Title
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    There's an alternative way if you're interested

    • one year ago
  14. hartnn Group Title
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    sure ! @jim_thompson5910

    • one year ago
  15. sedighn Group Title
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    hehe, yes please :)

    • one year ago
  16. jim_thompson5910 Group Title
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    it only works for square roots though

    • one year ago
  17. jim_thompson5910 Group Title
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    Alternative Way: Let z = a+bi z = a+bi z^2 = (a+bi)(a+bi) z^2 = a^2 + 2ab*i + bi^2 z^2 = a^2 + 2ab*i + b^2(-1) z^2 = a^2 + 2ab*i - b^2 z^2 = a^2 - b^2 + 2ab*i z^2 = (a^2 - b^2) + (2ab)*i So if z = a+bi, then z^2 = (a^2 - b^2) + (2ab)*i The real part of z^2 is a^2 - b^2 and the imaginary part of z^2 is 2ab ---------------------------------------------------------------------------------------- The idea is that if z = a+bi is a square root of some complex number w, then z^2 = w Now let w = 5+2i z^2 = w z^2 = 5+2i (a^2 - b^2) + (2ab)*i = 5+2i The real part of 5+2i is 5, so a^2 - b^2 = 5 The imaginary part of 5+2i is 2, so 2ab = 2 ---------------------------------------------------------------------------------------- You now have these two equations a^2 - b^2 = 5 2ab = 2 with 2 unknowns. You can use these two equations together (with the use of substitution) to solve for 'a' and 'b'

    • one year ago
  18. sedighn Group Title
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    I seee... thanks heaps

    • one year ago
  19. jim_thompson5910 Group Title
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    you're welcome

    • one year ago
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