arg(Z/i+1) when Z=-1+i When I think of this I just make it -1+i/i+1 but I dont know what to do next

Let's simplify -1+i/i+1   multiply numerator and denominator by i-1...so we have

[i - 1][i-1]/[(i+1)(i-1)] = (i^2 - 2i + 1)/(i^2 -1) = (-2i) / (-2) = i =  0 + 1i

tan(1/0) = undefined = pi/2

So...arg(z) = pi/2 + n*(2pi) for n = 0, ±1,±2, ±3.......

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

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

$$\frac{-i+1+i^2-i}{i^2-i+i-1}=\frac{-2i}{-2}$$   because i² = -1

$$\frac{-2i}{-2}=i$$

Now arg(i) = 1  because arg(x+iy) = √(x² + y²)

arg(i) = √(0² + (1)²) = 1

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