The answer is
[tex](-9+3\sqrt{2})+(3\sqrt{2}-2)i[/tex]
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Work Shown:
[tex](-3+\sqrt{-2})(3-\sqrt{2})[/tex]
[tex](-3+\sqrt{-1*2})(3-\sqrt{2})[/tex]
[tex](-3+\sqrt{-1}*\sqrt{2})(3-\sqrt{2})[/tex]
[tex](-3+i\sqrt{2})(3-\sqrt{2})[/tex]
[tex]x(3-\sqrt{2})[/tex] let [tex]x=-3+i\sqrt{2}[/tex]
[tex]3x-x*\sqrt{2}[/tex] distribute
[tex]3(x)-\sqrt{2}(x)[/tex]
[tex]3(-3+i\sqrt{2})-\sqrt{2}(-3+i\sqrt{2})[/tex]
[tex]-9+3i\sqrt{2}+3\sqrt{2}-i*\sqrt{2}*\sqrt{2}[/tex]
[tex]-9+3i\sqrt{2}+3\sqrt{2}-2i[/tex]
[tex]-9+3\sqrt{2}+3i\sqrt{2}-2i[/tex]
[tex](-9+3\sqrt{2})+(3\sqrt{2}-2)i[/tex]
The use of parenthesis in the last step is to help separate out the terms.
The last expression shown above is in the form a+bi where
[tex]a=-9+3\sqrt{2}[/tex]
[tex]b=3\sqrt{2}-2[/tex]
