Answer
$S=\{(\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15},\frac{3}{\sqrt 15}+\frac{2i}{\sqrt 15}),(\frac{2}{\sqrt 15}+\frac{3i}{\sqrt 15},\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15})\}$
Work Step by Step
We are given $v_1=(1-i,3+2i)\\
v_2=(2+3i,1-i)$
$v$ and $w$ are orthogonal if
$(v_1,v_2)=0 \\
= ((1-i,3+2i),(2+3i,1-i))\\
=(1-i)\bar {(2+3i)}+(3+2i)\bar {(1-i)} \\
=(1-i)(2-3i)+(3+2i)(1+i) \\
=0$
Hence, $v_1$ and $v_2$ are orthogonal vectors in $R^2$
To determine an orthogonal set, we obtain:
$$\frac{v_1}{||v_1||}=\frac{(1-i,3+2i)}{||(1-i,3+2i)||}=\frac{(1-i,3+2i)}{\sqrt ((1-i,3+2i),(1-i,3+2i))}=\frac{(1-i,3+2i)}{\sqrt 15}=(\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15},\frac{3}{\sqrt 15}+\frac{2i}{\sqrt 15})$$
$$\frac{v_2}{||v_2||}=\frac{(2+3i,1-i)}{||(2+3i,1-i)||}=\frac{(2+3i,1-i)}{\sqrt 15}=(\frac{2}{\sqrt 15}+\frac{3i}{\sqrt 15},\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15})$$
Hence, a corresponding orthonormal set of vector is
$S=\{(\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15},\frac{3}{\sqrt 15}+\frac{2i}{\sqrt 15}),(\frac{2}{\sqrt 15}+\frac{3i}{\sqrt 15},\frac{1}{\sqrt 15}-\frac{i}{\sqrt 15})\}$
