Write the complex number #(-5 - 3i)/(4i)# in standard form?

2 Answers
Mar 4, 2018

#(-5-3i)/(4i)=-3/4+5/4i#

Explanation:

We want the complex number in the form #a+bi#. This is a bit tricky because we have an imaginary part in the denominator, and we can't divide a real number by an imaginary number.

We can however solve this using a little trick. If we multiply both top and bottom by #i#, we can get a real number in the bottom:

#(-5-3i)/(4i)=(i(-5-3i))/(i*4i)=(-5i+3)/(-4)=-3/4+5/4i#

Mar 4, 2018

#-3/4+5/4i#

Explanation:

#color(orange)"Reminder"color(white)(x)i^2=(sqrt(-1))^2=-1#

#"multiply numerator/denominator by "4i#

#rArr(-5-3i)/(4i)xx(4i)/(4i)#

#=(-20i-12i^2)/(16i^2)#

#=(12-20i)/(-16)#

#=12/(-16)-(20i)/(-16)#

#=-3/4+5/4ilarrcolor(red)"in standard form"#