Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
1 & i &-2 \\ 3 & 4i & -5 \\ -1 & -3i & i
\end{bmatrix}$
Since $x,y,z \in nullspace (A)$ we obtain
$Ax=0\\ \begin{bmatrix}
1 & i &-2 \\ 3 & 4i & -5 \\ -1 & -3i & i
\end{bmatrix}\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}=0\\
\begin{bmatrix}
x +iy-2z \\ 3x+ 4iy -5 z\\ -1x -3iy+ iz
\end{bmatrix}=0\\
\rightarrow x +iy -2z =0\\
3x+4iy -5z =0\\
-x+ -3iy +iz =0\\
\rightarrow -4(x+iy-2z)+3x+4iy-5z=-x+3z=0 \rightarrow x=3z\\
\rightarrow x +iy -2z +(-x-3iy+iz)=-2iy-2z+iz=-2iy+z(i-2)=0 \rightarrow y=\frac{i-2}{2i}z$
Substitue $3z+i(\frac{i-2}{2i}z)-2z=0 \rightarrow \frac{iz}{2}=0 \rightarrow z=0\\
x=3z=3.0=0\\
y=\frac{i-2}{2i}.0=0$
Hence, nullspace $(A)= \{(0,0,0)\}$
