How do you convert $1 + j \sqrt{3}$ $1+jsqrt3$ to polar form?

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How do you convert $1 + j \sqrt{3}$ $1+jsqrt3$ to polar form?

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Answer 1 · 2016-07-23T06:40:23

$1 + j \sqrt{3} = 2 (cos (\frac{π}{3}) + j sin (\frac{π}{3}))$ $1+jsqrt3=2(cos(pi/3)+jsin(pi/3))$

Explanation:

To write the complex number, say $a + j b$ $a+jb$ , where $j^{2} = - 1$ $j^2=-1$ in polar coordinates, we write them as

$z = r (cos θ + j sin θ)$ $z=r(costheta+jsintheta)$ .

Now as $r cos θ = a$ $rcostheta=a$ and $r sin θ = b$ $rsintheta=b$ .

$r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ , $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^(-1)(b/a)$

Hence here for complex number $1 + j \sqrt{3}$ $1+jsqrt3$

$r = \sqrt{1^{2} + {(\sqrt{5})}^{2}} = \sqrt{4} = 2$ $r=sqrt(1^2+(sqrt5)^2)=sqrt4=2$ and $θ = {tan}^{- 1} (\frac{\sqrt{3}}{1}) = {tan}^{- 1} \sqrt{3} = \frac{π}{3}$ $theta=tan^(-1)(sqrt3/1)=tan^(-1)sqrt3=pi/3$

Hence $1 + j \sqrt{3} = 2 (cos (\frac{π}{3}) + j sin (\frac{π}{3}))$ $1+jsqrt3=2(cos(pi/3)+jsin(pi/3))$

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How do you convert $1 + j \sqrt{3}$ $1+jsqrt3$ to polar form?

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