Question 1009881
The multiplicative inverse of a number, {{{ a}}} is {{{1/a}}} so that

{{{a *(1/a) = 1}}}


so, the multiplicative inverse of {{{(1/3-i)}}}  is:

 {{{1/(1/3-i) }}}


 {{{1/(1/3-3i/3) }}}


 {{{1/((1-3i)/3) }}}


{{{3/(1-3i) }}}


{{{3(1+3i)/((1-3i)(1+3i)) }}}


{{{(3+9i)/(1^2-(3i)^2) }}}


{{{(3+9i)/(1-9i^2) }}}


{{{(3+9i)/(1-9(-1)) }}}


{{{(3+9i)/(1+9) }}}


{{{(3+9i)/10 }}}


{{{highlight(3/10+9i/10) }}}


check if 

{{{(1/3-i)(3/10+9i/10)=1}}}

{{{(3/10)(1/3)+(9i/10)(1/3)-i(3/10)-i(9i/10)=1}}}

{{{(3/10)(1/3)+(9i/10)(1/3)-(3i/10)-(9i^2/10)=1}}}

{{{(cross(3)1/10)(1/cross(3)1)+(cross(9)3i/10)(1/cross(3)1)-(3i/10)-(9i^2/10)=1}}}

{{{(1/10)+(3i/10)-(3i/10)-(9i^2/10)=1}}}

{{{(1/10)+cross((3i/10))-cross((3i/10))-(9(-1)/10)=1}}}

{{{(1/10)+9/10=1}}}

{{{10/10=1}}}

{{{1=1}}}