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Let $A=\begin{pmatrix} a & c \\ 0 & b \end{pmatrix}$ where $a \neq b $ or $c \neq 0$. $a,b,c $ being complex numbers.

I have to find every $X \in T_2^+(C) $ such that $X^2=A $.

I thing I have proved A admits square root(s) if and only if $a \neq 0$ and $b \neq 0$.

But then I can't figure out the square roots of A. I get the following equations: (If $X=\begin{pmatrix} d & f\\ 0 & e \end{pmatrix}$)

$d^2=a $

$e^2=b $

$(d+e)f=c $

But I don't know anything about square roots in C so I'm stuck..

Thanks!

angryavian
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Urefeu
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1 Answers1

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Write $a=r(\cos\alpha +i\sin\alpha)$ and then $d=r^{1/2}(\cos\alpha/2 +i\sin\alpha/2)$. Write $b=r(\cos\theta+i\sin\theta)$ and then $e=r^{1/2}(\cos\theta/2 +i\sin\theta/2)$. Finally after get $d$ and $e$ just replace at $f=c/(d+e)$ to get $f$.

Arnaldo
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  • Thanks, I forgot to use this notation. Yet, an other solution would be $r^{\frac{1}{2}}(cos( \frac{\alpha}{2} + \pi) + i \ sin(\frac{\alpha}{2}+\pi))$, and same for $e$ so we get four possible cases. Is there then a way to handle the case $d+e=0$ ie $|d|=|e|$ and $arg(d)=arg(e)$ without having to handle 4 different cases for the arguments? – Urefeu Dec 05 '16 at 08:15
  • If $d+e=0 \Rightarrow c=0$ and then $f$ can be anyone. That means you have infinite solutions for $X$. – Arnaldo Dec 05 '16 at 12:06
  • Yes it does but I have to find out all these solutions, that's why I'm talking about four cases – Urefeu Dec 05 '16 at 12:07
  • Well, for that last case you have $X=\begin{pmatrix} d & f\ 0 & -d \end{pmatrix}$ and $d$ has the form as the answer. – Arnaldo Dec 05 '16 at 12:12