To evaluate square roots of $1+2i$.

Question

Here is as far as I got.

First we write $1+2i$ in the polar form which is $\sqrt{5}e^{i\alpha}$ ($\alpha$ is the argument of $1+2i$ which turns out to be $\arctan2$). Therefore the square roots are $\pm \sqrt{\sqrt{5}} (\cos{\alpha/2}+i\sin{\alpha/2})$.

The answer given at the back is $\pm (\sqrt{\frac{\sqrt{5}+1}{2}}+i\sqrt{\frac{\sqrt{5}-1}{2}})$. How to I get it into this form?

See http://www.qc.edu.hk/math/Advanced%20Level/Finding%20the%20square%20root%20of%20a%20complex%20number.htm — lab bhattacharjee, Mar 27 '16 at 06:53
http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number — Hans Lundmark, Mar 27 '16 at 09:54

score 2 · Accepted Answer · answered Mar 27 '16 at 06:59

2

setting $$\sqrt{1+2i}=a+bi$$ we get the system $$1=a^2-b^2$$ and $$2=2ab$$

answered Mar 27 '16 at 06:59

Dr. Sonnhard Graubner

95,283

To evaluate square roots of $1+2i$.

1 Answers1