Question about time shifting property I know that time shifting property of discrete-time Fourier Transform states that 
$x[n]\leftrightarrow X(e^{j\omega})\implies x[n-n_0]\leftrightarrow e^{-j\omega n_0}X(e^{j\omega})$
So given a signal $\;y[n]=u[n]-u[n-5]\;,\;$What is $\;\mathfrak{F}\{ y_{(2)}[n-1]\}\:?$
$ y_{(2)}[n]\;$ denotes the slowed-down version of $\:y[n].\;$And $\:y[n] \leftrightarrow e^{-j2\omega}\frac{\sin(5\omega/2)}{\sin(\omega/2)}$
From time expansion property $\:y_{(2)}[n]\leftrightarrow e^{-j4\omega}\frac{\sin(5\omega)}{sin(\omega)}\:$then follows the time shifting property
$\mathfrak{F}\{ y_{(2)}[n-1]\}=e^{-j6w}\frac{\sin(5\omega)}{\sin(\omega)}$
But the answer should be $\:e^{-j5\omega}\frac{\sin(5w)}{\sin(\omega)}$
What is wrong with my understanding?
$y_{(2)}[n] = y[n/2]$ then $Y_{(2)}(j\omega) = 2 Y(j2\omega) = 2e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)}$.

$z[n] = y_{(2)}[n - 1]$ then $Z(j\omega) = e^{-j\omega}Y_{(2)}(j\omega) = e^{-j\omega}2e^{-j4\omega}\frac{\sin(5\omega)}{\sin(\omega)} = 2e^{-j5\omega}\frac{\sin(5\omega)}{\sin(\omega)}$.