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Roots of complex numbers

The question is find the 5th root of 1+2i. I have found r to be sqrt(5) but finding theta is causing me trouble because I'm getting a decimal and I'm not sure how to carry on with a decimal :frown:
1+2i can be written as √5*ei*arctan2

so if a 5th root of 1 + 2i is w ....

w5 = √5*ei*arctan2

...
Reply 2
Avatar for Zacken
Zacken
22
Original post by Lunu
The question is find the 5th root of 1+2i. I have found r to be sqrt(5) but finding theta is causing me trouble because I'm getting a decimal and I'm not sure how to carry on with a decimal :frown:


1+2i=5(cos⁡(arctan⁡2+2kπ)+isin⁡(arctan⁡2+2kπ))\displaystyle 1 + 2i = \sqrt{5} \left(\cos (\arctan 2 + 2k\pi) + i\sin (\arctan 2 + 2k\pi) \right) or you just leave it as a decimal and continue normally?

i.e: (1+2i)1/5=(5)1/5(cos⁡1.107+2kπ5+isin⁡1.107+2kπ5)\displaystyle (1 + 2i)^{1/5} = (\sqrt{5})^{1/5} \left(\cos \frac{1.107 + 2k\pi}{5} + i \sin \frac{1.107 + 2k\pi}{5}\right).

But I'd leave it as arctan⁡2\arctan 2 to be honest. Then it's just:

(1+2i)1/5=(5)1/5(cos⁡arctan⁡2+2kπ5+isin⁡arctan⁡+2kπ5)\displaystyle (1 + 2i)^{1/5} = (\sqrt{5})^{1/5} \left(\cos \frac{\arctan 2 + 2k\pi}{5} + i \sin \frac{\arctan + 2k\pi}{5}\right)
(edited 8 years ago)
Express a+bi in terms of r(cos(theta + 2kpi) +isin(theta + 2kpi)). To find the fifth root, every term is to the power of 1/5 (in the case of arguments they're divided by 5) and then sub in different whole values for k to find the argument. Remember the argument has to be within minus pi and positive pi.
Reply 4
Avatar for B_9710
B_9710
18
Original post by the bear
1+2i can be written as √5*ei*arctan2

so if a 5th root of 1 + 2i is w ....

w5 = √5*ei*arctan2

...


Do you prefer doing these questions with complex numbers in exponential form?
Original post by B_9710
Do you prefer doing these questions with complex numbers in exponential form?


i find it convenient... you can always do sin and cos later...
Reply 6
Avatar for Lunu
Lunu
OP
1
Thanks everyone! :smile:

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