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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
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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hartnn Group TitleBest ResponseYou've already chosen the best response.2
do you know how to convert the form a+bi into polar form, r <theta ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
and viceversa ?
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{5+2i}\] ?
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
Yes I do... I've never seen such a question so I'm not sure how to begin!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
convert 5+2i into polar first
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
okay so... \[\left[ \sqrt{29},0.381 \right]\]
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
(in radians)
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{r(\cos \theta +i \sin \theta)} = \sqrt{r}(\cos \frac{\theta}{2} +i \sin \frac{\theta}{2})\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
so just take square root of r half the angle and then convert it back to a+bi form
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
wow okay thanks a lot! we didn't do this method in class o.o
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
well, now you are smarter than your classmates ;) (Y)
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
There's an alternative way if you're interested
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
sure ! @jim_thompson5910
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
hehe, yes please :)
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
it only works for square roots though
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.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'
 one year ago

sedighn Group TitleBest ResponseYou've already chosen the best response.0
I seee... thanks heaps
 one year ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.1
you're welcome
 one year ago
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