How to I write the quotient #5/i# in standard form?

1 Answer
Nov 21, 2016

Answer:

#5/i=-5i#

Explanation:

In general, a complex number multiplied by its conjugate is real, as #(a+bi)(a-bi) = a^2-(bi)^2 = a^2+b^2#. Thus, to change an expression with a complex number for a denominator into one with a real denominator, one can multiply the numerator and the denominator by the conjugate of the denominator:

#z/(a+bi) = (z(a-bi))/((a+bi)(a-bi)) = (z(a-bi))/(a^2+b^2)#

In the given example, the conjugate of the denominator #i# is #bar(i) = -i#. Thus, we can multiply the numerator and the denominator by #-i# to simplify:

#5/i = (5*(-i))/(i*(-i))#

#=(5i)/(i^2)#

#=(5i)/(-1)#

#=-5i#