Comments on: C++ program to perform arithmetic operations of two complex numbers using operator overloading

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By: chill chilly kirubha

chill chilly kirubha — Wed, 06 Nov 2013 12:11:56 +0000

it’s easy to understand. thanks sir

By: chill chilly kirubha

chill chilly kirubha — Wed, 06 Nov 2013 12:11:10 +0000

it’s very simple and easy to understand. thanks sir

By: guru

guru — Fri, 15 Mar 2013 10:30:48 +0000

totally wrong

By: Robert L. Smith

Robert L. Smith — Thu, 06 Dec 2012 15:32:42 +0000

The above program for division of complex numbers will fail for very small or very large real or imaginary parts of the denominator because of overflow or underflow of the parts. About 50 years ago I showed a simple method that works in almost all cases where the result can be represented in a computer. The trick is to break the problem into two parts, depending on whether the absolute value of real part of the denominator is greater of less than the absolute value of the imaginary part of the denominator. Consider Q = (a + ib)/(c + id). Case 1: If abs(c) > than abs(d), then note that the denominator can be written (c + id) = c*(1 + ir) where r = d/c. Now instead of multiplying both the numerator and denominator by the complex conjugate of (c + id), just multiply both by the complex conjugate of (1 + ir). Note that the absolute value of r is less than 1. Note that the new denominator will become c*(1 + ir)*(1 – ir) = c*(1 + r*r) = c*(1 + r*d/c) = c + d*r). The quotient is:

q = (a + ib) / (c + id) = ((a + ib)*(1 – ir)) / (c*(1 + ir)*(1 – id/c))

= ((a + b*r) + i(b – a*r)) / (c*(1 + r*d/c))

= ((a + b*r)/(c + r*d)) + i((b – a*r)/(c + r*d))

Note that the numerator and denominator no longer involve squares of possibly large or small quantities.

Case 2: If abs(c) <= abs(d) then just reverse things a little bit so that the denominator (c + id) = d*(r + i) where r = c/d. In this case the absolute value of r is less than or equal to 1. In computing the quotient, we multiply the numerator and denominator by (r – i). The quotient can now be computed as follows:

q = (a + ib) / (c + id) = ((a + ib)*(r – i)) / (d*(r + i)*(c/d – i))

= ((a*r + b) + i(b*r – a)) / (c*r + d)

= ((a*r + b) / (d + c*r)) + i((-a + b*r) / (d + c*r))

I might mention that there are a few obscure cases where this method can be improved, again by breaking it down to two more sub-cases. My original paper appears in the Journal of the ACM.

By: aiswarya

aiswarya — Sat, 01 Sep 2012 07:36:35 +0000

too lengthy program