How do you simplify #(1/2+sqrt3i)/(1-sqrt2i)#?

1 Answer
Shwetank Mauria
Jun 11, 2017

#(1/2+sqrt3i)/(1-sqrt2i)=(1/(2sqrt2)-sqrt3)+(1/2+sqrt3/sqrt2)i#

Explanation:

To simpplify ratio of two complex numbers such as #(a+bi)/(c+di)#, just multiply numerator and denominator by the complex conjugate of denominator i.e. #c-di #.

Here, we have #(1/2+sqrt3i)/(1-sqrt2i)# and hence we should multipky numerator and denominator by #1+sqrt2i#

#:.(1/2+sqrt3i)/(1-sqrt2i)=((1/2+sqrt3i)(1+sqrt2i))/((1-sqrt2i)(1+sqrt2i))#

= #(1/2+(sqrt2i)/2+sqrt3i+sqrt6i^2)/(1^2-(sqrt2i)^2)#

= #(1/2+(sqrt2i)/2+sqrt3i+sqrt6(-1))/(1-2i^2)#

= #(1/2-sqrt6+(1/sqrt2i+sqrt3i))/(1-2(-1))#

= #(1-2sqrt6+(sqrt2+2sqrt3)i)/(2sqrt2)#

= #(1-2sqrt6)/(2sqrt2)+(sqrt2+2sqrt3)i/(2sqrt2)#

= #(1/(2sqrt2)-sqrt3)+(1/2+sqrt3/sqrt2)i#

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