How do you simplify #(-5-3i)/(4i)# and write the complex number in standard form?
1 Answer
Explanation:
To simplify, we require to multiply the numerator and denominator by the
#color(blue)"complex conjugate"# of the denominator.Given a complex number z = x ± yi then the complex conjugate is.
#color(red)(|bar(ul(color(white)(a/a)color(black)(barz=x∓yi)color(white)(a/a)|)))# Note that the real part remains unchanged while the
#color(blue)"sign"# of the imaginary part is reversed.Example: z = 5 + 4i then complex conjugate is
#barz=5-4i# Here z = 4i = 0 + 4i in standard form.Thus the complex conjugate is - 4i
Multiply numerator and denominator by - 4i
#rArr((-5-3i)color(red)"-4i")/(4ixxcolor(red)"-4i")=(20i+12i^2)/(-16i^2)#
#color(orange)"Reminder"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))#
#rArr(20i+12i^2)/(-16i^2)=(20i-12)/16=(-12)/16+20/16i#
#rArr(-5-3i)/(4i)=-3/4+5/4i" in standard form"#
