Can you solve Solve (z+5)/(z-i) = i (linear algebra)

The answer is:

z = -2 -2i

I have no idea how he ended up with this answer.

(z + 5) / (z - i) = i
z + 5 = i(z - i)
z + 5 = iz + 1
z - iz = -4
z(1 - i) = -4
z = -4/(1 - i)
z = -4(1 + i) / [(1 - i)(1 + i)]
z = -4(1 + i) / (1² - i²)
z = -4(1 + i) / 2
z = -2(1 + i)
z = -2 - 2i

Hello,

(z + 5) / (z - i) = i ← This means z≠i
z + 5 = i(z - i) ← Multiply both sides by (z-i)
z + 5 = iz + 1 ← Expand
z - iz = -4 ← Group zeds on one side, scalar on the other
z(1 - i) = -4 ← Factor
z = -4/(1 - i) ← Divide both side by (1 - i)
z = -4(1 + i) / [(1 - i)(1 + i)] ← Multiply top and bottom of fraction by the conjugate
z = -4(1 + i) / (1² - i²) ← Because (a+b)(a-b)=a²-b²
z = -4(1 + i) / 2 ← Because i²=-1
z = -2(1 + i)
z = -2 - 2i
QED.

Regards,
Dragon.Jade :-)

Assuming z-i cannot be 0, re-arrange like normal algebra

(z+5) / (z-i) = i

=> z+5=i*(z-i)
= i*z - i*i
= i*z + 1 {i=square root of -1}

=> (1-i)*z= - 4

=> z= - 4 / (1-i)
= - 4*(1+i) / [ (1-i) * (1+i) ]
= - 4*(1+i) / (1 - i^2)
= - 4*(1+i) / (1+1)
= - 4*(1+i) / 2
= -2* (1+i)

=> z=-2 -2i

z= -2 - 2 i