Answer
$\dfrac{2-3i}{-7i}=\dfrac{3}{7}+\dfrac{2}{7}i$
Work Step by Step
$\dfrac{2-3i}{-7i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{2-3i}{-7i}=\dfrac{2-3i}{-7i}\cdot\dfrac{7i}{7i}=\dfrac{7i(2-3i)}{-49i^{2}}=\dfrac{7(2i-3i^{2})}{-49i^{2}}=...$
Substitute $i^{2}$ with $-1$ and simplify:
$...=\dfrac{7[2i-3(-1)]}{-49(-1)}=\dfrac{7(2i+3)}{49}=\dfrac{2i+3}{7}=\dfrac{3}{7}+\dfrac{2}{7}i$
