How do you find the 4th root of 2(cos⁡(π4)+isin⁡(π4))?

The principal 4th root is: 2(cos⁡(π4)+isin⁡(π4))4=24(cos⁡(π16)+isin⁡(π16)) There are 3 other 4-th roots. Explanation: In general, if n is a positive integer and θ∈(−π,π] then the principal n-th root is given by: r(cos⁡θ+isin⁡θ)n=rn(cos⁡(θn)+isin⁡(θn)) So in our example: 2(cos⁡(π4)+isin⁡(π4))4=24(cos⁡(π16)+isin⁡(π16)) Note that the primitive Complex 4-th root of 1 is i, so the other 4-th roots are: i24(cos⁡(π16)+isin⁡(π16))=24(cos⁡(5π16)+isin⁡(5π16)) i224(cos⁡(π16)+isin⁡(π16))=24(cos⁡(−7π16)+isin⁡(−7π16)) i324(cos⁡(π16)+isin⁡(π16))=24(cos⁡(−3π16)+isin⁡(−3π16))

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Jaeden Frederick

Answered question

2022-02-01

How do you find the 4th root of

2 (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))

Answer & Explanation

Telering3b

Beginner2022-02-02Added 11 answers

The principal 4th root is:

\sqrt[4]{2 (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))} = \sqrt[4]{2} (\cos (\frac{π}{16}) + i \sin (\frac{π}{16}))

There are 3 other 4-th roots. Explanation: In general, if n is a positive integer and

θ \in (- π, π]

then the principal n-th root is given by:

\sqrt[n]{r (\cos θ + i \sin θ)} = \sqrt[n]{r} (\cos (\frac{θ}{n}) + i \sin (\frac{θ}{n}))

So in our example:

\sqrt[4]{2 (\cos (\frac{π}{4}) + i \sin (\frac{π}{4}))} = \sqrt[4]{2} (\cos (\frac{π}{16}) + i \sin (\frac{π}{16}))

Note that the primitive Complex 4-th root of 1 is i, so the other 4-th roots are:

i \sqrt[4]{2} (\cos (\frac{π}{16}) + i \sin (\frac{π}{16})) = \sqrt[4]{2} (\cos (\frac{5 π}{16}) + i \sin (\frac{5 π}{16}))

i^{2} \sqrt[4]{2} (\cos (\frac{π}{16}) + i \sin (\frac{π}{16})) = \sqrt[4]{2} (\cos (- \frac{7 π}{16}) + i \sin (- \frac{7 π}{16}))

i^{3} \sqrt[4]{2} (\cos (\frac{π}{16}) + i \sin (\frac{π}{16})) = \sqrt[4]{2} (\cos (- \frac{3 π}{16}) + i \sin (- \frac{3 π}{16}))

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