- #1
pinsky
- 96
- 0
I have a problem in understanding the procedures of a solved example. It goes like this.
[tex]\left ( \frac{z+i}{z-i} \right )^4 = -1[/tex]
Therefor we can write:
[tex]\left | \frac{z+i}{z-i} \right | = 1[/tex]
From that we can see that z is a real number because:
[tex]\left | z+i \right | = \left | z-i \right | \; \; \; , z=x+yi \; \; \Rightarrow y=0 \; \; z=x[/tex]
So let's say that:
[tex]W = \frac{x+i}{x-i}=\frac{x^2-1}{x+1}+\frac{2x}{x+1}i[/tex]
[tex]W^4 = -1 [/tex]
And from that they get the solutions by doing some additional steps which i understand. What i can't seem to understand is the transformation written in red. For I'm having a homework to solve:
[tex]\left ( \frac{z+1}{z-1} \right )^3 = -1[/tex]
and I don't know if i can apply the same assumption as in the upper example cause i don't understand what happened.
Any help is appreciated.
[tex]\left ( \frac{z+i}{z-i} \right )^4 = -1[/tex]
Therefor we can write:
[tex]\left | \frac{z+i}{z-i} \right | = 1[/tex]
From that we can see that z is a real number because:
[tex]\left | z+i \right | = \left | z-i \right | \; \; \; , z=x+yi \; \; \Rightarrow y=0 \; \; z=x[/tex]
So let's say that:
[tex]W = \frac{x+i}{x-i}=\frac{x^2-1}{x+1}+\frac{2x}{x+1}i[/tex]
[tex]W^4 = -1 [/tex]
And from that they get the solutions by doing some additional steps which i understand. What i can't seem to understand is the transformation written in red. For I'm having a homework to solve:
[tex]\left ( \frac{z+1}{z-1} \right )^3 = -1[/tex]
and I don't know if i can apply the same assumption as in the upper example cause i don't understand what happened.
Any help is appreciated.