Question

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

Answer 1

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

Answer 2

and vice-versa ?

Answer 3

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

Answer 4

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

Answer 5

convert 5+2i into polar first

Answer 6

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

Answer 7

(in radians)

Answer 8

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

Answer 9

yes ^

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

There's an alternative way if you're interested

Answer 14

sure ! @jim_thompson5910

Answer 15

hehe, yes please :)

Answer 16

it only works for square roots though

Answer 17

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


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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


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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'

Answer 18

I seee... thanks heaps

Answer 19

you're welcome