How do you convert $1 + 4 i$ $1 + 4i$ to polar form?

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How do you convert $1 + 4 i$ $1 + 4i$ to polar form?

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Answer 1 · 2016-05-15T03:37:29

$≅ \sqrt{17} (cos 1.33 + i sin 1.33)$ $~=sqrt(17)(cos1.33 + isin1.33)$

Explanation:

If $z = a + b i$ $z = a +bi$

Then $z$ $z$ may be expressed in polar form as:

$z = r (cos θ + i sin θ)$ $z = r(cos theta + i sin theta)$
Where: $r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2 + b^2)$ and $θ = arctan (\frac{b}{a})$ $theta = arctan(b/a)$

In this example: $z = 1 + 4 i$ $z=1+4i$
Hence: $a = 1$ $a=1$ and $b = 4$ $b=4$

Therefore: $r = \sqrt{1^{2} + 4^{2}}$ $r = sqrt(1^2+4^2)$ = $\sqrt{17}$ $sqrt(17)$
And: $θ = arctan (\frac{4}{1})$ $theta = arctan(4/1)$ $≅ 1.33$ $~= 1.33$

Hence: $z ≅ \sqrt{17} (cos 1.33 + i sin 1.33)$ $z ~= sqrt(17)(cos 1.33 + i sin 1.33)$

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How do you convert $1 + 4 i$ $1 + 4i$ to polar form?

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