We are asked to write the following fraction of complex numbers in standard form: [tex]\displaystyle{ \frac{3-i}{1+5i} [/tex].
Whenever we are dividing by a complex number by another complex number a+bi, to write the result in standard form, we multiply the expressions in the numerator and denominator by the conjugate of a+bi, that is a-bi:
[tex]\displaystyle{ \frac{3-i}{1+5i}= \frac{(3-i)(1-5i)}{(1+5i)(1-5i)} [/tex]
in the numerator we distribute (3-i) to 1 and -5i, in the denominator we use the difference of squares formula [tex](a-b)(a+b)=a^2-b^2[/tex], and [tex]i^2=-1[/tex]:
[tex]\displaystyle{ \frac{(3-i)(1-5i)}{(1+5i)(1-5i)}= \frac{(3-i)\cdot1+(3-i)\cdot(-5i)}{1^2-(5i)^2}= \frac{3-i-15i-5}{1-25i^2}[/tex]
[tex]\displaystyle{= \frac{-2-16i}{1+25}= \frac{-2-16i}{26}=-1/13-(8/13)i [/tex]
Answer: D
