How do you simplify #(1-isqrt2)/(1+isqrt2)#?

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Answer 1 · 2018-04-16T02:43:21

#=-1/3(1+2sqrt2i)#

Expression #=(1-isqrt2)/(1+isqrt2)#

Rationalise the denominator by multiplying the expression by #(1-isqrt2)/(1-isqrt2)#

Expression #= (1-isqrt2)/(1+isqrt2) xx (1-isqrt2)/(1-isqrt2)#

#= (1-isqrt2)^2/(1^2-(isqrt2)^2#

#= (1-2sqrt2i+(sqrt2i)^2)/(1^2-(sqrt2i)^2#

Remember: #i^2 =-1#

#:.# Expression #= (1-2sqrt2i-2)/(1-(-2))#

#= (-1-2sqrt2i)/3#

#=-1/3(1+2sqrt2i)#