Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

sedighn

  • 2 years ago

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

  • This Question is Closed
  1. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    do you know how to convert the form a+bi into polar form, r <theta ?

  2. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and vice-versa ?

  3. dumbcow
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sqrt{5+2i}\] ?

  4. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes I do... I've never seen such a question so I'm not sure how to begin!

  5. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    convert 5+2i into polar first

  6. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay so... \[\left[ \sqrt{29},0.381 \right]\]

  7. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (in radians)

  8. dumbcow
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\sqrt{r(\cos \theta +i \sin \theta)} = \sqrt{r}(\cos \frac{\theta}{2} +i \sin \frac{\theta}{2})\]

  9. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes ^

  10. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    so just take square root of r half the angle and then convert it back to a+bi form

  11. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow okay thanks a lot! we didn't do this method in class o.o

  12. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    well, now you are smarter than your classmates ;) (Y)

  13. jim_thompson5910
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    There's an alternative way if you're interested

  14. hartnn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    sure ! @jim_thompson5910

  15. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hehe, yes please :)

  16. jim_thompson5910
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it only works for square roots though

  17. jim_thompson5910
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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'

  18. sedighn
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I seee... thanks heaps

  19. jim_thompson5910
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you're welcome

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.