Question #0ae21

1 Answer
Oct 12, 2016

Answer:

#(1+i)/(i(2+3i))=-1/13-5/13i#

Explanation:

Given a complex number #z=a+bi#, the complex conjugate of #z#, denoted #bar(z)#, is given by #bar(z)=a-bi#. A useful property of the complex conjugate is that #zbar(z) = a^2+b^2 in RR#. We can use this to eliminate the complex value from the denominator of the expression.

#(1+i)/(i(2+3i)) = (1+i)/(2i+3i^2)#

#=(1+i)/(-3+2i)#

#=((1+i)(-3-2i))/((-3+2i)(-3-2i))#

#=(-3-3i-2i-2i^2)/((-3)^2+2^2)#

#=(-1-5i)/13#

#=-1/13-5/13i#