Complex Numbers Question 10

In the complex numbers, where $i^2=-1$ , $\dfrac{2-i}{-3+i}$ = ?

-\dfrac{2}{3} - i

-\dfrac{5}{8}+\dfrac{1}{8}i

-\dfrac{7}{8}+\dfrac{1}{8}i

-\dfrac{5}{10}+\dfrac{1}{10}i

-\dfrac{7}{10}+\dfrac{1}{10}i

Approach

We need to multiply by the conjugate of the denominator to rationalize all square roots.

\frac{2-i}{-3+i}

\frac{2-i}{-3+i}\cdot \frac{-3-i}{-3-i}

= \frac{(2-i)(-3-i)}{(-3+i)(-3-i)}

= \frac{-6+3i-2i+i^2}{9-i^2}

= \frac{-6+i+(-1)}{9-(-1)}

= \frac{-7+i}{10}

= \boxed{-\frac{7}{10}+\frac{1}{10}i}