How do you simplify #1/(2+3i)# and write in a+bi form?
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Multiply by the complex conjugate.
#frac1(2+3i)(frac(2-3i)(2-3i))=frac(2-3i)(4-6i+6i-9i^2)=frac(2-3i)(4-9i^2)#
Recall that #i^2=-1# since #i=sqrt(-1)#.
#=frac(2-3i)(4-(-9))=(2-3i)/13=2/13-3/13i#
#1# may be written in rectangular form as #1+0i#, and in polar form as #1/_0^@#.
Similarly, #2+3i# can be written in polar form as
#sqrt(2^2+3^2)/_tan^(-1)(3/2) =sqrt13/_56,31^@#
We may now divide them by using the standard rule for division of complex numbers in polar form :
#(1/_0)/( sqrt13/_56,31^@)=1/(sqrt13)/_(0-56,31)#
This answer may now be converted back to rectangular form using standard trig relationships to get #(1/sqrt13)cos(-56,31)+i(1/sqrt13)sin(-56,31)#
#=0,1538-0,23078i#
(Note : We could have maintained complex rectangular form and multiply top and bottom by the complex conjugate of the denominator #bar z = 2-3i# and then multiply top and bottom. Bottom would then be a real number, and top would require rule of multiplication of complex numbers in rectangular form - quite a complex rule ;) - Hence probably quicker the way I did it by rather converting everything to polar form first)
