How do you divide #(1+9i)/(-2+9i)#?

1 Answer
Jul 27, 2016

Answer:

#79/85-27/85i#.

Explanation:

Let #z=(1+9i)/(-2+9i)#

Then usual Method to simplify this is to multiply both #Nr.# and #Dr.# by the conjugate of the complex no. of the #Dr.#, i.e., to say,

#z=(1+9i)/(-2+9i) xx (-2-9i)/(-2-9i)#

#=(-2-9i-18i-81i^2)/{-2^2-(9i)^2}#

#=(-2-27i-81(-1))/(4-81i^2)#

#=(79-27i)/85#

#=79/85-27/85i#.

But, we can, as well, solve it as follows :-

Suppose that, #z=x+iy=(1+9i)/(-2+9i)#

#rArr (x+iy)(-2+9i)=1+9i#

#rArr -2x+9ix-2iy+9i^2y=1+9i#

#rArr -2x+i(9x-2y)+9(-1)y=1+9i#

#rArr (-2x-9y)+i(9x-2y)=1+9i#

Comparing the Real and Imaginary Parts of both sides, we get,

#(1) : 2x+9y=-1, and, (2) : 9x-2y=9#

#:. 2xx(1)+9xx(2) rArr (4x+18y)+(81x-18y)=-2+81#

#rArr 85x=79# #rArr x=79/85#.

Using this with #(2)#, we have,

#2y=9(x-1)=9(79/85-1)=9(-6/85)#

#rArr y=-(9*6)/(85*2)=-27/85#, giving,

#z=x+iy=79/85-27/85i#, as before!

Enjoy Maths!