Brock Brown

2022-01-18

How do you divide $\frac{-2-5i}{3i}$ ?

Navreaiw

Beginner2022-01-19Added 34 answers

To solve $\frac{-2-5i}{3i}$

multiply numerator and denominator by i (note${i}^{2}=-1$ )

and we get$\frac{i\cdot (-2-5i)}{i\cdot \left(3i\right)}\text{}\text{or}\text{}\frac{-2i-5{i}^{2}}{3{i}^{2}}$

or$\frac{-2i-5(-1)}{3\cdot -1}\text{}i.e.\text{}\frac{5-2i}{-3}$ , which is equivalent to

Answer:$-\frac{5}{3}+\left(\frac{2}{3}\right)i$

multiply numerator and denominator by i (note

and we get

or

Answer:

Hattie Schaeffer

Beginner2022-01-20Added 37 answers

The denominator of the fraction is required to be real. To achieve this multiply the numerator and denominator by 3i.

$\frac{-2-5i}{3i}\times \frac{3i}{3i}$

$=\frac{-6i-15{i}^{2}}{9{i}^{2}}$

[Note:${i}^{2}={\left(\sqrt{-1}\right)}^{2}=-1$ ]

$=\frac{-6i+15}{-9}=\frac{2}{3}i-\frac{5}{3}$

[Note:

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