Question #6d4d4

1 Answer
Jim G.
Jul 7, 2017

#3+i" or " -3-i#

Explanation:

#"let " sqrt(8+6i)=a+bilarr" square both sides"#

#rArr8+6i=(a+bi)^2=a^2-b^2+2abi#

#"equating the coefficients of real/imaginary parts"#

#a^2-b^2=8to(1)#

#2ab=6rArrb=3/ato(2)#

#"substitute " b=3/a" in " (1)#

#rArra^2-9/a^2=8#

#"multiply through by "a^2#

#rArra^4-9=8a^2#

#rArra^4-8a^2-9=0#

#rArr(a^2-9)(a^2+1)=0#

#a^2+1rArra^2=-1" has no real solutions"#

#a^2-9rArra^2=9rArra=+-3#

#"substituting in " (2)#

#rArrb=+-1#

#rArrsqrt(8+6i)=3+i" or " -3-i#

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