Prove that

Given: $z=2-3 i$

To Prove: $z^{2}-4 z+13=0$

Taking LHS, $z^{2}-4 z+13$

Putting the value of $z=2-3 i$, we get

$(2-3 i)^{2}-4(2-3 i)+13$

$=4+9 i^{2}-12 i-8+12 i+13$

$=9(-1)+9$

$=-9+9$

$=0$

$=\mathrm{RHS}$

Hence, $z^{2}-4 z+13=0$

Now, we have to deduce $4 z^{3}-3 z^{2}+169$

Now, we will expand $4 z^{3}-3 z^{2}+169$ in this way so that we can use the above equation i.e. $z^{2}-4 z+13$

$=4 z^{3}-16 z^{2}+13 z^{2}+52 z-52 z+169$

Re – arrange the terms,

$=4 z^{3}-16 z^{2}+52 z+13 z^{2}-52 z+169$

$=4 z\left(z^{2}-4 z+13\right)+13\left(z^{2}-4 z+13\right)$

$=4 z(0)+13(0)[$ from eq. (i) $]$

$=0$

$=\mathrm{RHS}$

Hence Proved