What is $frac{z_1+z_2}{z_1−z_2}$ if $z_1,z_2inmathbb{C}$, such that $z_1neq z_2,|z_1|neq|z_2|$












1












$begingroup$



Let $z_1$ and $z_2$ be two complex numbers such that $z_1≠z_2$ and $|z_1|neq|z_2|$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $frac{z_1+z_2}{z_1−z_2}$ may be ___________



a) zero or purely imaginary



b) real and +ve



c) real and -ve




My reference gives the solution "zero or purely imaginary". If it was $|z_1|=|z_2|$ I can easily find the solution from geometry as the complex numbers $z_1$ and $z_2$ form a rhombus and $z_1+z_2$ and $z_1-z_2$ be its diagonals and they intersect orthoganally, thus $argBig(frac{z_1+z_2}{z_1-z_2}Big)=arg(z_1+z_2)-arg(z_1-z_2)=pmfrac{pi}{2}$, thus purely imaginary. But, in this case how do I find the solution ?






proof-verification complex-numbers







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  • 1




    $begingroup$
    For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
    $endgroup$
    – Martin R
    Jan 31 at 10:31










  • $begingroup$
    @MartinR My reference gives that solution, i'm confused.
    $endgroup$
    – ss1729
    Jan 31 at 10:34






  • 1




    $begingroup$
    Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
    $endgroup$
    – Reinhard Meier
    Jan 31 at 11:20










  • $begingroup$
    I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
    $endgroup$
    – user376343
    Jan 31 at 22:38


















1












$begingroup$



Let $z_1$ and $z_2$ be two complex numbers such that $z_1≠z_2$ and $|z_1|neq|z_2|$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $frac{z_1+z_2}{z_1−z_2}$ may be ___________



a) zero or purely imaginary



b) real and +ve



c) real and -ve




My reference gives the solution "zero or purely imaginary". If it was $|z_1|=|z_2|$ I can easily find the solution from geometry as the complex numbers $z_1$ and $z_2$ form a rhombus and $z_1+z_2$ and $z_1-z_2$ be its diagonals and they intersect orthoganally, thus $argBig(frac{z_1+z_2}{z_1-z_2}Big)=arg(z_1+z_2)-arg(z_1-z_2)=pmfrac{pi}{2}$, thus purely imaginary. But, in this case how do I find the solution ?






proof-verification complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
    $endgroup$
    – Martin R
    Jan 31 at 10:31










  • $begingroup$
    @MartinR My reference gives that solution, i'm confused.
    $endgroup$
    – ss1729
    Jan 31 at 10:34






  • 1




    $begingroup$
    Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
    $endgroup$
    – Reinhard Meier
    Jan 31 at 11:20










  • $begingroup$
    I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
    $endgroup$
    – user376343
    Jan 31 at 22:38
















1












1








1





$begingroup$



Let $z_1$ and $z_2$ be two complex numbers such that $z_1≠z_2$ and $|z_1|neq|z_2|$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $frac{z_1+z_2}{z_1−z_2}$ may be ___________



a) zero or purely imaginary



b) real and +ve



c) real and -ve




My reference gives the solution "zero or purely imaginary". If it was $|z_1|=|z_2|$ I can easily find the solution from geometry as the complex numbers $z_1$ and $z_2$ form a rhombus and $z_1+z_2$ and $z_1-z_2$ be its diagonals and they intersect orthoganally, thus $argBig(frac{z_1+z_2}{z_1-z_2}Big)=arg(z_1+z_2)-arg(z_1-z_2)=pmfrac{pi}{2}$, thus purely imaginary. But, in this case how do I find the solution ?






proof-verification complex-numbers







share|cite|improve this question











$endgroup$





Let $z_1$ and $z_2$ be two complex numbers such that $z_1≠z_2$ and $|z_1|neq|z_2|$. If $z_1$ has positive real part and $z_2$ has negative imaginary part, then $frac{z_1+z_2}{z_1−z_2}$ may be ___________



a) zero or purely imaginary



b) real and +ve



c) real and -ve




My reference gives the solution "zero or purely imaginary". If it was $|z_1|=|z_2|$ I can easily find the solution from geometry as the complex numbers $z_1$ and $z_2$ form a rhombus and $z_1+z_2$ and $z_1-z_2$ be its diagonals and they intersect orthoganally, thus $argBig(frac{z_1+z_2}{z_1-z_2}Big)=arg(z_1+z_2)-arg(z_1-z_2)=pmfrac{pi}{2}$, thus purely imaginary. But, in this case how do I find the solution ?






proof-verification complex-numbers




proof-verification complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 31 at 10:41







ss1729

















asked Jan 31 at 10:25









ss1729ss1729

2,05411124




2,05411124








  • 1




    $begingroup$
    For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
    $endgroup$
    – Martin R
    Jan 31 at 10:31










  • $begingroup$
    @MartinR My reference gives that solution, i'm confused.
    $endgroup$
    – ss1729
    Jan 31 at 10:34






  • 1




    $begingroup$
    Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
    $endgroup$
    – Reinhard Meier
    Jan 31 at 11:20










  • $begingroup$
    I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
    $endgroup$
    – user376343
    Jan 31 at 22:38
















  • 1




    $begingroup$
    For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
    $endgroup$
    – Martin R
    Jan 31 at 10:31










  • $begingroup$
    @MartinR My reference gives that solution, i'm confused.
    $endgroup$
    – ss1729
    Jan 31 at 10:34






  • 1




    $begingroup$
    Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
    $endgroup$
    – Reinhard Meier
    Jan 31 at 11:20










  • $begingroup$
    I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
    $endgroup$
    – user376343
    Jan 31 at 22:38










1




1




$begingroup$
For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
$endgroup$
– Martin R
Jan 31 at 10:31




$begingroup$
For $z_1 = 2$ and $z_2 = -i$ I get $(z_1+z_2)/(z_1 - z_2) = (3+4 i)/5$, therefore I cannot see how "zero or purely imaginary" can be the correct solution.
$endgroup$
– Martin R
Jan 31 at 10:31












$begingroup$
@MartinR My reference gives that solution, i'm confused.
$endgroup$
– ss1729
Jan 31 at 10:34




$begingroup$
@MartinR My reference gives that solution, i'm confused.
$endgroup$
– ss1729
Jan 31 at 10:34




1




1




$begingroup$
Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
$endgroup$
– Reinhard Meier
Jan 31 at 11:20




$begingroup$
Are you sure that it is not $z_1neq z_2$ and $|z_1|=|z_2|$?
$endgroup$
– Reinhard Meier
Jan 31 at 11:20












$begingroup$
I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
$endgroup$
– user376343
Jan 31 at 22:38






$begingroup$
I agree with Reinhard Meier, there is apparently a typo. If $|z_1|neq|z_2|,;$ all three possibilities can occur.
$endgroup$
– user376343
Jan 31 at 22:38












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