Question

I need help with this practice problem solving It asks to make sure to divide and put your answer in standard form for complex numbers

Answer 1

$-\frac{17}{26}-\frac{7}{26}i$

Explanation

$\frac{2-3i}{-1+5i}$

Step 1

Multiply both, numerator and denominator by the conjugate of the denominator and expand

so

$\begin{gathered} \frac{2-3i}{-1+5i} \\ \frac{2-3i}{-1+5i}\cdot\frac{-1-5i}{-1-5i} \\ \frac{2-3i}{-1+5i}\cdot\frac{-1-5i}{-1-5i}=\frac{(2-3i)(-1-5i)}{(-1+5i)(-1-5i)} \\ \frac{2-3i}{-1+5i}\cdot1=\frac{(2-3i)(-1-5i)}{(-1+5i)(-1-5i)} \\ \frac{2-3i}{-1+5i}=\frac{-2-10i+3i+15i^2}{(-1)^2-(5i)^2} \\ \frac{2-3i}{-1+5i}=\frac{-2-7i+15(-1)}{(1)-(25i^2)^{}} \\ \frac{2-3i}{-1+5i}=\frac{-2-7i-15}{(1)-(25)(-1)^{}} \\ \frac{2-3i}{-1+5i}=\frac{-2-7i-15}{(1)+25^{}}=\frac{-17-7i}{26^{}} \\ \frac{2-3i}{-1+5i}=\frac{-17-7i}{26^{}} \\ \end{gathered}$

now, get the standar d form z=a+bi

$\begin{gathered} \frac{2-3i}{-1+5i}=\frac{-17-7i}{26^{}} \\ \frac{2-3i}{-1+5i}=-\frac{17}{26}-\frac{7}{26}i \end{gathered}$

so,the answer is

$-\frac{17}{26}-\frac{7}{26}i$

I hope this helps you