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    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
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    1+2i can be written as √5*ei*arctan2

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

    w5 = √5*ei*arctan2

    ...
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    (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
    \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: \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 to be honest. Then it's just:

    \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)
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    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.
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    (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?
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    (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...
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    Thanks everyone!
 
 
 
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