Question 1b

SCH_CALC L3CALC

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$$z=2-\frac{2+i}{iz}$$

Multiply by $iz$:

$$i{z}^2=2iz-(2+i)⟺i{z}^2-2iz+(2+i)$$

Using quadratic formula:

$$a=i,b=-2i,c=2+i$$ $$z=\frac{2i\pm \sqrt{(-2i{)}^2-4i(2+i)}}{2i}$$ $$z=\frac{2i\pm \sqrt{-4-8i+4}}{2i}$$ $$z=\frac{2i\pm \sqrt{-8i}}{2i}$$ $$z=\frac{2i\pm (2-2i)}{2i}$$

Hence the solutions for $z$ are:

$$z=\frac{2i+2-2i}{2i}⟹\frac{1}{i}$$ $$z=\frac{2i-2+2i}{2i}⟹\frac{4i-2}{2i}⟹\frac{2i-1}{i}$$

But.. JUST IN CASE, we rationalize, and we get:

$$z=\frac{1}{i}⟹-i$$ $$z=\frac{-i(2i-1)}{-i^2}⟹2+i$$

$\boxed{z=-i,z=2+i}$