# complex number equation

• Nov 8th 2012, 12:33 PM
jakobjakob
complex number equation
I need to solve the following equation and i am encountering some problem

Re: (z+i)/(1-z)=0

This is what i did so far:
Re:
(z+i)/(1-z) +1 = 1
(1+i)/(z-1)=-1
.
and then i multiply left side with (z(conjugate)-1)/(z(conjugate)-1).
After all the calculations i get the following equation and i don't know which a or b value is correct:
a(a+1)+b(b+1)=0

z=complex number, a= Re value of complex number, b=Im value of complex number

I hope i've expressed myself clear and thank you for your help.(Happy)
• Nov 8th 2012, 12:42 PM
MarkFL
Re: complex number equation
For the given equation, this implies $\displaystyle z+i=0$.
• Nov 8th 2012, 02:58 PM
Plato
Re: complex number equation
Quote:

Originally Posted by jakobjakob
I need to solve the following equation and i am encountering some problem
Re: (z+i)/(1-z)=0

I think the actual problem is $\displaystyle \text{Re}\left(\frac{z+i}{1-z}\right)=0$

Note that $\displaystyle \text{Re}\left(\frac{z}{w}\right)=\frac{\text{Re}( z)\text{Re}(w)+\text{Im}(z)\text{Im}(w)}{|w|^2}$

If you want that to be 0, then the numerator must be zero.

Also $\displaystyle z=-i~\&~0$ are two solutions.
• Nov 8th 2012, 04:53 PM
Prove It
Re: complex number equation
How could z = -i possibly be a solution to this equation? The real parts of a complex number are always real...

\displaystyle \displaystyle \begin{align*} \textrm{Re}\left( \frac{z + i}{1 - z} \right) &= 0 \\ \textrm{Re}\left[ \frac{x + y\,i + i}{1 - \left( x + y\,i \right)} \right] &= 0 \\ \textrm{Re}\left[ \frac{x + \left( y + 1 \right) i}{1 - x - y\,i} \right] &= 0 \\ \textrm{Re}\left\{ \frac{\left[ x + \left( y+ 1\right) i \right]\left( 1 - x + y\,i \right)}{\left( 1 - x - y\, i \right) \left( 1 - x + y\, i\right) } \right\} &= 0 \\ \textrm{Re}\left[ \frac{x(1 - x) + xy\,i + (1 - x)\left( y + 1 \right) i + y\left( y + 1 \right) i^2}{(1 - x)^2 + y^2 } \right] &= 0 \\ \textrm{Re}\left\{ \frac{x(1 - x) - y(y + 1) + \left[ xy + (1 - x)(y + 1) \right] i }{(1 - x)^2 + y^2} \right\} &= 0 \\ \textrm{Re} \left\{ \frac{x(1 - x) - y(y + 1)}{(1 - x)^2 + y^2} + \left[\frac{xy + (1 - x)(y + 1)}{(1 - x)^2 + y^2}\right] i \right\} &= 0 \end{align*}

\displaystyle \displaystyle \begin{align*} \frac{x(1 - x) - y(y + 1)}{(1 - x)^2 + y^2} &= 0 \\ x(1 - x) - y(y + 1) &= 0 \textrm{ for all }(x, y) \neq (1, 0) \\ x - x^2 - y^2 - y &= 0 \\ x^2 - x + y^2 + y &= 0 \\ x^2 - x + \left( -\frac{1}{2} \right)^2 + y^2 + y + \left( \frac{1}{2} \right)^2 &= \left( -\frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 \\ \left( x - \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 &= \left( \frac{\sqrt{2}}{2} \right)^2 \end{align*}

So the solution to your problem is all points that lie on the circle of radius \displaystyle \displaystyle \begin{align*} \frac{\sqrt{2}}{2} \end{align*} units, centred at \displaystyle \displaystyle \begin{align*} \left( \frac{1}{2}, -\frac{1}{2} \right) \end{align*}.