# Complex number

While learning about numbers we come across Real Numbers and Imaginary Numbers. Real numbers consist of the whole numbers, rational numbers and the irrational numbers. Imaginary numbers are the numbers which are not real. A combination of a real number and an imaginary number together is called a complex number and is denoted by a+ib, where a is a real part, b is an imaginary part and ‘i’ stands for square root of (-1). The word complex is similar to a building complex in which all the buildings are joined, here a real number and an imaginary number are joined to get a complex number. Some of the Complex Numbers Problems can be given as follows, add (4+3i) + (2+5i), the real part and the imaginary parts are added separately and the sum written, (4+2) + (3+5)i= 6+8i. Subtract (7+3i) – (3+2i) = (7-3)+ (3-2)i= 4-I, multiply (3+4i)x(3-4i) this is in the form (a+b)x(a-b)=a^2-b^2 and hence we get, 3^2 – (4i)^2 = 3-(-16) = 3+16 = 19.

While we solve complex numbers with different operations we follow a series of steps. In a Complex Numbers Tutorial the steps shown are: Addition operation: (a+bi) + (c+di) = (a+b) + (c+d)iHere the real part and the imaginary parts are grouped and added separately, the sum is again written in complex number form. (6+3i) + (2+7i)=(6+2) + (3+7)i= 8 + 10is

Subtraction operation: (a+bi) – (c+di) = (a-c) + (c-d)i; (-12+5i) – (3 – 7i) = (-12-3) + (5 -7)i= -15 – 2i

Multiplication operation: (a+bi) x (c+di) = a(c+di) + bi(c+di)= ac +adi +bci +bd(i)^2 we have i="sqrt(-1)" and hence i^2=-1, substituting this value and regrouping, (ac-bd)+(ad+bc)i. Here the FOIL method is used to multiply the terms.

Before learning to solve using division operation we shall learn about conjugate complex numbers. Conjugate here means to change the operation from addition to subtraction or vice versa. The conjugate of (a+bi) would be (a-bi) and the conjugate of (a-bi) would be (a+bi). It is represented by a bar over the complex number. Conjugate complex numbers help in the division of complex numbers. Example: Divide (3-5i)/(4-7i). The conjugate of (4-7i) is (4+7i). Multiply the numerator and denominator with the conjugate, (3-5i)(4+7i)/(4-7i)(4+7i). The complex numbers in the numerator are multiplied using the FOIL method and the complex numbers in the denominator are multiplied using the identity (a+b)(a-b)=a^2-b^2. [12 + 21i – 20i -35(i^2)]/(16-(49 i^2); we have i^2= -1 substituting this value and simplifying we get,[12 + i – 35(-1)]/16-(-49)] = [12+35+i]/16+49 = (47+i)/65. The answer should be in the a+bi form, (47/65) + (1/65)i