How do you simplify #(sqrt3-i)div(2-2sqrt3i)#?

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Answer 1 · 2016-12-04T14:46:08

The answer is #=sqrt3/4+i/4#

When you have a fraction of complex numbers,

#w=z_1/z_2#

Multiply numerator and denominator by the conjugate of the denominator

#w=(z_1*barz_2)/(z_2*barz_2)#

If the complex number is #z=a+ib#

The conjugate is #barz=a-ib#

and #i^2=-1#

#w=(sqrt3-i)/(2-2sqrt3i)#

#=((sqrt3-i)(2+2sqrt3i))/((2-2sqrt3i)(2+2sqrt3i))#

#=(2sqrt3+2*3i-2i-2sqrt3i^2)/(4-4*3i^2#

#=(4sqrt3+4i)/16#

#=sqrt3/4+i/4#