<s>[INST] Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$. 
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.

I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
 [/INST] You're gonna need to convert $\sqrt {2-i}$ to the form $\alpha + i \beta$.  
To do that, convert to polar
$$ 2-i=r(\cos \theta + i \sin \theta)$$
So that
$$\sqrt{2-i}=\sqrt r \cdot (\cos \theta + i \sin \theta)^{1/2} = \sqrt r \cdot \left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right) $$
Some of the ingredients are:
$$r=\sqrt{2^2+(-1)^2}=\sqrt 5, \quad \cos \theta= \frac{2}{\sqrt 5}, \quad \sin \theta =\frac{-1}{\sqrt 5} $$
$\theta$ will be a negative angle, so that 
$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\cos \theta}{2}} \qquad \sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\cos \theta}{2}}$$
Continuing
$$\cos{\frac{\theta}{2}}=\sqrt{\frac{1+\frac{2}{\sqrt 5}}{2}}= \sqrt{\frac{\sqrt 5+2}{2\sqrt 5}}=\frac{\sqrt{2+\sqrt 5}}{\sqrt 2 \sqrt r}$$
$$\sin{\frac{\theta}{2}}=-\sqrt{\frac{1-\frac{2}{\sqrt 5}}{2}}= -\sqrt{\frac{\sqrt 5-2}{2\sqrt 5}}=-\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 \sqrt r}$$
Making
$$\sqrt{2-i}=\left(\frac{\sqrt{2+\sqrt 5}}{\sqrt 2}\right)-i \left(\frac{\sqrt{-2+\sqrt 5}}{\sqrt 2 }\right) \in \alpha + i \beta$$
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