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z=a+bi (a+bi)(3+2i)=3(a+bi)+3+2i (3a-2b)+(2a+3b)i=(3a+3)+(3b+2)i therefore 3a-2b=3a+3 2a+3b=3b+2 b=-3/2,a=1 So z = 1-3i / 2
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