What does #(1-3i)/sqrt(1+3i)# equal?

1 Answer
George C.
Jan 5, 2016

#(1-3i)/sqrt(1+3i)#

#=(-2sqrt((sqrt(10)+1)/2)+3/2sqrt((sqrt(10)-1)/2))-(2sqrt((sqrt(10)-1)/2)+3/2sqrt((sqrt(10)+1)/2))i#

Explanation:

In general the square roots of #a+bi# are:

#+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)#

See: http://socratic.org/questions/how-do-you-find-the-square-root-of-an-imaginary-number-of-the-form-a-bi

In the case of #1+3i#, both Real and imaginary parts are positive, so it's in Q1 and has a well defined principal square root:

#sqrt(1+3i)#

#=sqrt((sqrt(1^2+3^2)+1)/2)+sqrt((sqrt(1^2+3^2)-1)/2)i#

#=sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i#

So:

#(1-3i)/sqrt(1+3i)#

#=((1-3i)sqrt(1+3i))/(1+3i)#

#=((1-3i)^2 sqrt(1+3i))/((1+3i)(1-3i))#

#=((1-3i)^2 sqrt(1+3i))/4#

#=1/4(1-3i)^2 (sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)#

#=1/4(-8-6i)(sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)#

#=-1/2(4+3i)(sqrt((sqrt(10)+1)/2) + sqrt((sqrt(10)-1)/2)i)#

#=-1/2((4sqrt((sqrt(10)+1)/2)-3sqrt((sqrt(10)-1)/2))+(4sqrt((sqrt(10)-1)/2)+3sqrt((sqrt(10)+1)/2))i)#

#=(-2sqrt((sqrt(10)+1)/2)+3/2sqrt((sqrt(10)-1)/2))-(2sqrt((sqrt(10)-1)/2)+3/2sqrt((sqrt(10)+1)/2))i#

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