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

`-(17)/(26)-(7)/(26)i`

Step-by-step explanation

`(2-3i)/(-1+5i)`

Step 1

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

so

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

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

`\begin{gathered} (2-3i)/(-1+5i)=\frac{-17-7i}{26^{}} \\ (2-3i)/(-1+5i)=-(17)/(26)-(7)/(26)i \end{gathered}`

so,the answer is

`-(17)/(26)-(7)/(26)i`

I hope this helps you