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How the differences of the two squares a2-b2(a b)(a-b) relates to the product of a complex number and it's conjugate?
Wiki User
∙ 7y ago
(a+b)(a-b) = a2 - b2
if we replace b by ib in this equation, then we get:
(a+ib)(a-ib) = a2 - (ib)2 = a2 + b2
if z=a+ib then its conjuguate is a-ib and their product is a2 + b2
Wiki User
∙ 7y ago
Wiki User
∙ 8y agoIn real number, when multiplying the difference of two squares, the middle term disappears:(a + b)*(a - b) = a^2 - ab + ab + b^2 = a^2 - b^2
Similarly, with the complex number and its conjugate, the imaginary term cancels out.
(a + ib)*(a - ib) = a^2 - iab + iab + (ib)^2 = a^2 - i^2b^2 = a^2 + b^2.
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How the differences of the two squares relates to the product of a complex number and its conjugate?
In real number, when multiplying the difference of two squares, the middle term disappears:(a + b)*(a - b) = a^2 - ab + ab + b^2 = a^2 - b^2Similarly, with the complex number and its conjugate, the imaginary term cancels out.(a + ib)*(a - ib) = a^2 - iab + iab + (ib)^2 = a^2 - i^2b^2 = a^2 + b^2.
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