How do you simplify #(3+ isqrt2)/ (7-isqrt2)#?

1 Answer
Jan 30, 2016

Answer:

Multiplying denominator and numerator by the complex conjugate of the numerator, #(7+isqrt2)#, shows that this expression can be simplified to: #(21+10isqrt2-2)/(51)#

Explanation:

The complex conjugate of a number #(a+bi)# is #(a-bi)#, and multiplying top (denominator) and bottom (numerator) of a complex fraction by the complex conjugate of the bottom (numerator) will simpifly it:

#(3+isqrt2)/(7-isqrt2)*(7+isqrt2)/(7+isqrt2)#

Note that any number, including a complex number, divided by itself is 1, so in multiplying by #(7+isqrt2)/(7+isqrt2)# we are in effect multiplying by 1, leaving the result unchanged.

Note: #i*i=-1# and #sqrt2*sqrt2=2#

#(3+isqrt2)/(7-isqrt2)*(7+isqrt2)/(7+isqrt2)# = #(21+3isqrt2+7isqrt2-2)/(49+7isqrt2-7isqrt2+2)#
= #(21+10isqrt2-2)/(49+2) = (21+10isqrt2-2)/(51)#