If $z=1+i$, then the argument of ${ z }^{ 2 }{ e }^{ z-i }$ is

Let $w={ z }^{ 2 }{ e }^{ z-i }$

Put $z=1+i$

$\Rightarrow w={ \left( 1+i \right) }^{ 2 }{ e }^{ 1+i-i }=\left( 1+2i+{ i }^{ 2 } \right) e$

$=\left( 1+2i-1 \right) e=2ei$

Now, argument of $w=arg\left( w \right)$

$=\tan ^{ -1 }{ \left( \dfrac { 2e }{ 0 } \right) } =\tan ^{ -1 }{ \infty } =\dfrac { \pi }{ 2 }$