How do I write the complex number $1+i\sqrt{3}$ In polar form with argument $θ$ between $0$ and $2\pi$.

Question

How do I write the complex number $1+i\sqrt{3}$ In polar form with argument $θ$ between $0$ and $2\pi$?

This question is so hard I’ve asked everyone and they can’t figte it out!! I hope you guys can help me out with this problem!! This math class is killing me they won’t explain anything and I’m just sitting there all clueless. I hope this problem will be able to be solved!!! Pleaseeeee help

Answer 1

If $x$ and $y$ are positive real numbers (are in the first quadrant) then the argument of the complex number $z=x+yi$ is given by $$\operatorname{Arg}(z)=\arctan \left( \tfrac{y}{x} \right) $$

Answer 2

$\arctan (\sqrt 3)=\frac {\pi} 3$ so the answer is $2e^{i \frac {\pi} 3}$.

Answer 3

$$1+\sqrt3i=2\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)=2(\cos60^{\circ}+i\sin60^{\circ}).$$

Answer 4

Every complex number $z=x+iy$ can be written in the form of $z=re^{i\theta}$ where $$\theta=\operatorname{arg}(z)=\arctan \left( \tfrac{y}{x} \right)$$ and $$r=|z|=\sqrt{x^2+y^2}$$ So applying these 2 properties to change your variables from $(x,y)$ to $(r,\theta)$ you'll find in your example that $\theta=60^{\circ}=\frac{\pi}{3}$ and $r=2$

So $z=2e^{\frac{\pi}{3}i}$