Tuesday, 12 August 2008

Basic Complex Number Operations

Complex number does has the same set of basic math operations as normal algebra. The basic operations are the familiar addition, subtraction, multiplication and division.

The expression for complex number however differs in that it is splitted into 2 parts, namely, the REAL and IMAGINARY parts.

Format: Z = a + ib
where "a" represents the REAL part and "ib" the IMAGINARY part.

Let me move on with explanation of the basic math operations below.

Addition of Complex Number

Z1 = a + ib, and Z2 = c + id

If Z1 + Z2, then (a + ib ) + (c + id).

However, we need to separate the real parts addition from the imaginary parts addition.

Rewriting the above addition of Z1 and Z2, a + c + ib + id ==> (a +c) + i(b + d) answer.

NOTE: Take care and always be aware that complex number has real & imaginary terms. With that solving complex number is sweet and simple.

Subtraction of Complex Number

If we now perform Z1 - Z2, then the math operation yields (a + ib) - (c + id) .

The answer would then be, after taking care of the 2 different terms, a - c + ib - id.

Final answer : Z1 - Z2 = (a - c) + i(b - d).

Common mistake: Never take care of the sign for id when subtracting after opening brackets. The mistake is a - c + ib + id <== the wrong "+" sign of "id". The principle of basic algebra (math operations) still applies for Complex Number computation. Multiplication of Complex Number

If Z1 x Z2, then (a + ib)(c + id).

Do not be afraid!
The steps are the same as for normal algebra except that we need to take care of the "i" terms.

(a + ib)(c + id) = ac + iad + ibc + i2bd

Since i2 = -1, the working becomes ac + iad + ibc - bd.

We can now see that the last term became a REAL term! Take note of that when performing Complex Number multiplication.

Rewriting ac + iad + ibc - bd into real and imaginary terms,
==> (ad - bd) + i(ad + bc) answer.

The math operations of above examples can be seen to be simple, am I correct?
It is really simple! Not a bit complex at all.

How about Division of Complex Number?
Let me break this for a later post. I need to delay explanation of Division operation till I cover a special part in complex number computation which is the "Conjugate" concept.

:) And Hang On .....


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