## sedighn Group Title Find the square roots of 5+2i, giving your answers in the form a+bi one year ago one year ago

1. hartnn Group Title

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

2. hartnn Group Title

and vice-versa ?

3. dumbcow Group Title

$\sqrt{5+2i}$ ?

4. sedighn Group Title

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

5. hartnn Group Title

convert 5+2i into polar first

6. sedighn Group Title

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

7. sedighn Group Title

8. dumbcow Group Title

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

9. hartnn Group Title

yes ^

10. hartnn Group Title

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

11. sedighn Group Title

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

12. hartnn Group Title

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

13. jim_thompson5910 Group Title

There's an alternative way if you're interested

14. hartnn Group Title

sure ! @jim_thompson5910

15. sedighn Group Title

16. jim_thompson5910 Group Title

it only works for square roots though

17. jim_thompson5910 Group Title

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 Group Title

I seee... thanks heaps

19. jim_thompson5910 Group Title

you're welcome