Simplify $w=\frac{(1+i)z-i+1}{iz-1}$

Question

I have difficulties understanding how this expression $$w=\frac{(1+i)z-i+1}{iz-1}$$

is simplified to this $$w=1-i-2\cdot\frac{1+i}{z+i}$$

Here are some steps from my exercise notebook: $$\frac{1+i}{i}=1-i$$

I understand why is this done, it's what stands in numerator with $z$ over what stands with $z$ in denominator. $$w=\frac{(1+i)z\color{green}{-i+1}}{iz-1}=\frac{(1-i)(iz-1)\color{red}{+1-i}\color{green}{-i+1}}{iz-1}=1-i+\frac{2-2i}{iz-1}=\\=1-i+\frac{-2-2i}{z+i}$$

The only problem I have is this red marked expression. I don't know where that comes from, unlike green which is obvious from the first one.

Any help would be appreciated. I am not pro at math and I often have problems with simple things. Thank you very much!

Answer 1

The expression $1-i$ that you have in red is the difference $$ (1+i)z-(1-i)(iz-1)=z+iz-(iz-1+z+i)=1-i. $$ So what happened was insteading of writing $A\color{green}{-i+1}$, your exercist notebook wrote $B+1-i\color{green}{-i+1}$ where $$ A=(1+i)z,\quad B=(1-i)(iz-1),\quad\text{and, as we have just seen, } 1-i=A-B. $$

Answer 2

The solution is just skipping steps / being clever. Just take it step by step and you'll get the answer. Expanding the original numerator,

$$ w = \frac{(1+i)z -i +1}{iz -1} = \frac{z+iz -i + 1}{iz-1}$$

We want the denominator to be $z+i$, so divide by $i$ (which is the same as multiplying by $-i$) in the numerator and denominator to get

$$ w = \frac{z+iz -i + 1}{iz-1} \frac{-i}{-i} = \frac{-iz+z +1-i}{z+i}$$

Now we add and subtract what we need in the numerator to factor (making sure not to change the value):

$$ w = \frac{-i(z+\color{green}{i}) + (z + \color{green}{i}) +1-i +\color{red}{(-1-i)}}{z+i}$$

Now divide and we're done:

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