Searchingthe polar form of $z = 2+i(1+\sqrt{3})$

Question

Find the polar form of:

$$z = 2+i(1+\sqrt{3})$$

Is there any other way to solve that in different way than by finding $|z|$ that is really "ugly" ($|z| = \sqrt{8+2\sqrt{3}}$)?

Answer 1

you are right $z=\rho \cdot e^{i\cdot \theta}$ with $\rho=|z| = \sqrt{8+2\sqrt{3}}$. But Note that $$ \frac{1 + \sqrt3}{2} = \cos\frac{π}{3}+\sin\frac{π}{3}=\sqrt2\cos(\frac{π}{3}-\frac{π}{4} )=\sqrt2\cos\frac{π}{12}$$

and then ,

$$\theta=\arg (z)=\arctan\left(\frac{1 + \sqrt3}{2}\right)=\color{red}{\arctan\left(\sqrt2\cos\frac{π}{12}\right)}$$

Answer 2

For finding polar form of any complex number, you can't refrain from calculating its magnitude.

$$x+iy= r (\cos \theta + i \sin \theta) =r e^{i \theta}$$

You'll always need to find $r$ and $\theta $.

Answer 3

Yes it is correct $z=\rho \cdot e^{i\cdot \theta}$ with

$$\rho=|z| = \sqrt{8+2\sqrt{3}}$$

and

$$\theta=\arg (z)=\arctan\left(\frac{1 + \sqrt3}{2}\right)$$