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

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

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2convert 5+2i into polar first

sedighn
 2 years ago
Best ResponseYou've already chosen the best response.0okay so... \[\left[ \sqrt{29},0.381 \right]\]

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

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

sedighn
 2 years ago
Best ResponseYou've already chosen the best response.0wow okay thanks a lot! we didn't do this method in class o.o

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

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1There's an alternative way if you're interested

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.2sure ! @jim_thompson5910

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1it only works for square roots though

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1Alternative 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'

jim_thompson5910
 2 years ago
Best ResponseYou've already chosen the best response.1you're welcome
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