Question 27318: Can you help me find the multiplicative inverse?8-2i
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website!
Can you help me find the multiplicative inverse?8-2i
As the complex number (8-2i)is not (the additive identity comlex number)0 + 0i
it must have multiplicative inverse in the field of complex numbers.
The multiplicative inverse of (8-2i) is given by 1/(8-2i)
And 1/(8-2i) = 1/(8-2i) X (8+2i)/(8+2i)
(multiplying nr and dr by the complex conjugate (8+2i) of (8-2i)
That is equal to (8+2i))/[(8-2i)(8+2i))] = ((8+2i))/(8^2 + 2^2)
=(8+2i)/(64 + 4) = (8+2i)/68 =(8/68) + (2/68)i = (2/17) + (1/34)i
Note: The product of a non zero complex number (a+ib) and its conjuagate (a-ib) is real and is given by a^2 + b^2
as you may see it as follows:
(a+ib)(a-ib) = a^2 - (ib)^2 by actual multiplication of complex numbers
= a^2 - (i^2b^2) (using (mn)^2 = m^2n^2 )
= a^2 - (-b^2) (as i^2 = -1 by definition)
= a^2 + b^2
Therefore (a+ib)(a-ib) = a^2 + b^2
Note: If (a+ib) is any non zero complex number, then its multiplicative inverse is given by the formula: (a-ib)/(a^2+b^2)
and similarly if (a-ib) is any non zero complex number, then its multiplicative inverse is (a+ib)/(a^2+b^2)
And in the above problem (a-ib) is (8-2i)[where a = 8 and b = 2] and hence its multiplicative inverse is by formula given by (8+2i)/(8^2+2^2)=(8+2i)/68
In the above problem you may leave the answer as just (8+2i)/68 instead of simplifying and expressing it in the form (x+iy)
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