How do you convert $\sqrt{3 + 4 i}$ $sqrt(3+4i)$ to polar form?

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How do you convert $\sqrt{3 + 4 i}$ $sqrt(3+4i)$ to polar form?

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Answer 1 · 2017-01-12T05:34:35

$\pm \sqrt{5} c i s ({26.565}^{o})$ $+-sqrt5 cis(26.565^o)$ , nearly. See explanation.

Explanation:

Only for real $a^{2}$ $a^2$ , sqrta^2)=|a| #, by convention ( or definition ).

So, the other square root $- | a |$ $-|a|$ is out of reach.

Here, for sqrt(z), where z is complex, we cannot conveniently take

one and keep off the other.

So, $\sqrt{3 + 4 i} = {(3 + i 4)}^{\frac{1}{2}}$ $sqrt(3+4i)=(3+i4)^(1/2)$ that has two values.

They are $\sqrt{5} {(cos a + i sin a)}^{\frac{1}{2}}$ $sqrt5(cos a+isina)^(1/2)$ ,

where $cos a = \frac{3}{5} and sin a = \frac{4}{5}$ $cos a =3/5 and sin a = 4/5$

$= \sqrt{5} {(cos (a + 2 k π) + i sin (a + 2 k π))}^{\frac{1}{2}}, k = 0, 1$ $=sqrt5(cos(a+2kpi)+isin(a+2kpi))^(1/2), k =0, 1$

$\sqrt{5} (cos (\frac{a}{2} + k π) + i sin (\frac{a}{2} + k π)), k = 0, 1$ $sqrt5(cos(a/2+kpi)+isin(a/2+kpi)), k =0, 1$ ,

using De Moivre's theorem

$= \sqrt{5} (cos (\frac{a}{2}) + i sin (\frac{a}{2})) and \sqrt{5} (cos (\frac{a}{2} + π) + i sin (\frac{a}{2} + π))$ $=sqrt5(cos (a/2)+isin(a/2)) and sqrt5(cos (a/2+pi)+isin(a/2+pi))$

$= \pm \sqrt{5} (cos (\frac{a}{2}) + i sin (\frac{a}{2}))$ $=+-sqrt5(cos (a/2)+isin(a/2))$

$= \sqrt{5} c i s ({26.565}^{o})$ $=sqrt5cis(26.565^o)$ , nearly.

using $cos (π + θ) = - cos θ and sin (π + θ) = - sin θ$ $cos (pi+theta)=-cos theta and sin (pi+theta)=-sin theta$ .

Answer 2 · 2017-03-12T18:44:00

$\sqrt{3 + 4 i} = 2 + i = (\sqrt{5}, {tan}^{- 1} (\frac{1}{2}))$ $sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))$

Explanation:

Note that:

${(2 + i)}^{2} = 4 + 4 i + i^{2} = 3 + 4 i$ $(2+i)^2 = 4+4i+i^2 = 3+4i$

Since $3 + 4 i$ $3+4i$ is in Q1, $2 + i$ $2+i$ is its principal square root.

Further note that:

$| 2 + i | = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$ $abs(2+i) = sqrt(2^2+1^2) = sqrt(5)$

So we have:

$\sqrt{3 + 4 i} = 2 + i = (\sqrt{5}, {tan}^{- 1} (\frac{1}{2}))$ $sqrt(3+4i) = 2+i = (sqrt(5), tan^(-1)(1/2))$

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