anonymous
  • anonymous
Find the square roots of 5+2i, giving your answers in the form a+bi
Mathematics
jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
do you know how to convert the form a+bi into polar form, r
hartnn
  • hartnn
and vice-versa ?
dumbcow
  • dumbcow
\[\sqrt{5+2i}\] ?

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anonymous
  • anonymous
Yes I do... I've never seen such a question so I'm not sure how to begin!
hartnn
  • hartnn
convert 5+2i into polar first
anonymous
  • anonymous
okay so... \[\left[ \sqrt{29},0.381 \right]\]
anonymous
  • anonymous
(in radians)
dumbcow
  • dumbcow
\[\sqrt{r(\cos \theta +i \sin \theta)} = \sqrt{r}(\cos \frac{\theta}{2} +i \sin \frac{\theta}{2})\]
hartnn
  • hartnn
yes ^
hartnn
  • hartnn
so just take square root of r half the angle and then convert it back to a+bi form
anonymous
  • anonymous
wow okay thanks a lot! we didn't do this method in class o.o
hartnn
  • hartnn
well, now you are smarter than your classmates ;) (Y)
jim_thompson5910
  • jim_thompson5910
There's an alternative way if you're interested
hartnn
  • hartnn
anonymous
  • anonymous
hehe, yes please :)
jim_thompson5910
  • jim_thompson5910
it only works for square roots though
jim_thompson5910
  • jim_thompson5910
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'
anonymous
  • anonymous
I seee... thanks heaps
jim_thompson5910
  • jim_thompson5910
you're welcome

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