Solve an equation with complex numbers The question is to solve the following equation for complex numbers
$$z-i = iz +5$$
I have tried to add i to both sides which gives $$z = iz +5 + i$$
I have also tried with combinining all the terms on the LHS so i get $$z - i - iz - 5 = 0$$
Can you help with solving this equation?
 A: $$z- iz=5 +i$$
$$z=\frac{5+i}{1-i}$$
$$z=\frac{5+i}{1-i}\frac{1+i}{1+i}$$
$$z=2+3i$$
A: The fact that this equation is in complex numbers shouldn't give you problems. It is an equation of degree 1 of the type
$$az+b=0 \ \ \ \ \mbox{ with } a,b \in \Bbb{C},  a \neq 0$$
so the solution is simply
$$z= -\frac{b}{a}$$
In your particular case your equation is
$$(1-i)z + (-5-i)=0$$
so the solution is
$$z= \frac{5+i}{1-i}$$
Now, I hope you are able to compute the inverse of $1-i$, using conjugation
$$\frac{1}{1-i} = \frac{\overline{1-i}}{|1-i|^2} = \frac{1}{2}(1+i)$$
and so on
A: I remember when I first studied algebra, I had trouble with equations like:
$$3x  = x + 8.$$
It's hard to 'get $x$ on its own' when there are multiple $x$'s!
The trick, of course, is that
$$3x - x = 2x$$
so we can rearrange the above equation to get
$$3x-x = x + 8 - x$$
$$2x = 8 $$
$$x = 4$$
Your situation has complex numbers, but apart from that it's identical:
$$z - i = iz + 5$$
$$z - i + (i - iz) = iz + 5 + (i - iz)$$
$$z - iz = 5 + i$$
$$z(1-i) = 5 + i$$
$$z = \frac{5+i}{1-i}$$
$$z = \frac{5+i}{1-i}\frac{i+1}{i+1}$$
$$z = 2 + 3i.$$
A: Notice, we have $$z-i=iz+5$$ Let, $z=x+iy$ then we have $$x+iy-i=i(x+iy)+5$$
$$x+i(y-1)=ix+i^2y+5$$ setting $i^2=-1$ $$x+i(y-1)=ix-y+5$$
$$(x+y-5)+i(y-x-1)=0$$ Now, comparing the corresponding Real & imaginary parts of both the sides, we get $$x+y-5=0\tag 1 $$  $$y-x-1=0\tag 2$$ Now, on solving (1) & (2), we get $$x=2, y=3$$ Hence, we have 
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{z=2+3i}}$$
A: $$z-i = iz +5\Longleftrightarrow$$
$$z-i -(iz-i)= iz +5-(iz-i)\Longleftrightarrow$$
$$(1-i)z= 5+i\Longleftrightarrow$$
$$\frac{(1-i)z}{1-i}= \frac{5+i}{1-i}\Longleftrightarrow$$
$$z= 2+3i$$
