Am I going about this wrong? Complex expression to polar form

Question

I have the expression below, which I'd like to write in polar form. $$z = \frac{i}{{1+\frac{i(\sqrt3-1)}{1+i}}}$$

Own process

My process was very tedious; and I also wouldn't solve the final part without a calculator.

$$\frac{\sqrt3-1}{1+i} = \frac{(\sqrt3-1)(1-i)}{(1+i)(1-i)} = \frac{\sqrt3-i\cdot\sqrt{3} - 1 + i}{2}$$

$$\begin{align}\therefore z = \frac{i}{1+i\cdot\frac{\sqrt3-i\cdot\sqrt{3} - 1 + i}{2}} &= \frac{i}{\frac{2}{2} + \frac{i\cdot\sqrt3+\sqrt3-i-1}{2}}\\ &= \frac{i}{\frac{1}{2}+\frac{\sqrt3}{2}+i(\frac{\sqrt3}{2}-\frac{1}{2})} \end{align}$$

I then extend, once again, with the denominator's conjugate and get the following:

$$z = \frac{1}{4}(\sqrt3-1)+i\cdot\frac{1}{4}(1+\sqrt3)$$

This gives me that $|z| = \frac{1}{\sqrt2}$ after a little tedious work.

Now's the step that I can't solve without a calculator, as I don't know what angle whose cosinus would equal $$\frac{\Re z}{|z|} = \frac{\frac{1}{4}(\sqrt3-1)}{\frac{1}{\sqrt2}}$$

I suspect there is an easier way solving this, I'm not sure how though.

Edit 1. Added my attempt to find an angle $\theta$ which satisfies the equation in polar form.

$$\cos\theta = \frac{\Re z}{|z|} = \frac{\frac{1}{4}(\sqrt3-1)}{\frac{1}{\sqrt2}} = \frac{\sqrt3-1}{2\cdot\sqrt2}$$

How do I proceed from here? The final form would be $$z = \frac{1}{\sqrt2}(\cos\theta+i\cdot\sin\theta)$$

Answer 1

We have

$$z=\color{red}{\frac{\frac{2}{\sqrt2}\frac{\sqrt2}2}{\frac12\frac21}}\frac{-1+i}{1+i\sqrt3}=\frac1{\sqrt2}e^{3i\pi/4}e^{-i\pi/3}$$

Answer 2

$$1+\frac{i(\sqrt3-1)}{1+i}=\frac{1+i\sqrt3}{1+i}$$

$$\implies\frac i{1+\dfrac{i(\sqrt3-1)}{1+i}}=\frac{i(i+1)}{1+i\sqrt3}=\frac{(i-1)(1-i\sqrt3)}{1^2+(\sqrt3)^2}$$

$$=\frac{(\sqrt3+1)i+\sqrt3-1}4$$

Can you take it home from here?