How do you convert $1+jsqrt3$ to polar form?

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Answer 1 · 2016-07-23T06:40:23

$1+jsqrt3=2(cos(pi/3)+jsin(pi/3))$

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

$z=r(costheta+jsintheta)$ .

Now as $rcostheta=a$ and $rsintheta=b$ .

$r=sqrt(a^2+b^2)$ , $theta=tan^(-1)(b/a)$

Hence here for complex number $1+jsqrt3$

$r=sqrt(1^2+(sqrt5)^2)=sqrt4=2$ and $theta=tan^(-1)(sqrt3/1)=tan^(-1)sqrt3=pi/3$

Hence $1+jsqrt3=2(cos(pi/3)+jsin(pi/3))$

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