Question

1 + i supra 2 - 3i + 1 - i supra 2 + 3i

Answer 1

$\dfrac{(1+i)(2+3i)+(1-i)(2-3i)}{2^2-9i^2}=$

$=\dfrac{2+3i+2i-3+2-3i-2i-3}{7}=-\dfrac{2}{7}$
Am tinut sema ca i²=-1