First slide
Algebra of complex numbers
Question

If x+iy=a+ibc+id, then x2+y22c2+d2 equal 

Moderate
Solution

(x+iy)2(c+id)=a+ib(x+iy)2|c+id|=|a+ib||x+iy|2|c+id|=|a+ib|x2+y2c2+d2=a2+b2

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