A real number set contains irrational and rational numbers.

An imaginary set contains numbers which when squared gives a negative value.**For example:** $(\sqrt{-1})^{2}$ = -1.

i = $\sqrt{-1}$

A combination of real and imaginary numbers is called a complex number. It is denoted by ‘z’. **For example:** 2 + i, 6i + 5, 7i, 5.

This
implies that all real numbers are contained in the set of complex
numbers, as we can write any real number, say 6, as 6 + 0i.**Properties of ‘i’:**

1. i = $\sqrt{-1}$

2. i^{2} = -1

3. i^{3} = -i

4. i^{2n} = 1; where n is any natural number.**Adding and Subtracting Complex Numbers:**

To
add/subtract complex numbers, the real parts are added/subtracted and
the imaginary parts are added/subtracted and the result is obtained.**Example:** (3 + 2i) + (5 – i)**Solution:** 3 + 2i + 5 – i = (3 + 5) + (2 – 1) i = 8 + i**Multiplying Complex Numbers:**

Let (a + bi) and (c + di) be two complex numbers. Then the product of the two will be given by:

(a + bi)(c + di) = ac + bci + adi + bdi^{2} = ac + bci + adi - bd = (ac - bd) + i(bc + ad)**Example:** Solve (3 + 2i) (5 – i)**Solution:** (3 + 2i) (5 – i)

= 15 + 10i – 3i – 2i^{2}

= 15 + 7i – 2(-1) [i^{2} = -1]

=15 + 7i + 2 = 17 + 7i**Conjugates:**

If ‘a + bi’ is a complex number then ‘a – bi’ is its conjugate. It can be written as bar over it also.**Multiplying Conjugates:**

(a + bi)(a – bi) = a^{2} – (bi)^{2} = a^{2} – b^{2}i^{2} = a^{2} + b^{2}

So
when two conjugates are multiplied the result is the sum of the squares
of the real part and the coefficient of imaginary part. **Division in Complex Numbers:**

Conjugates are used in dividing complex numbers. We explain it with the help of an example.**Examp**l**e:** $\frac{2+3i}{4-5i}$**Solution:** $\frac{2+3i}{4-5i}$

=$\frac{(2+3i)(4+5i)}{(4-5i)(4+5i)}$

=$\frac{8+12i+10i-15}{16-25i^{2}}$

=$\frac{22i-7}{16+25}$

=$\frac{-7+22i}{41}$

=$\frac{-7}{41}+\frac{22i}{41}$

Some more example problems on complex numbers:**Example 1:** Simplify $\sqrt{-25}$**Solution:** $\sqrt{-25}$ = $\sqrt{-1 \times 25}$ = $\sqrt{-1} \times \sqrt{25}$ = $5\sqrt{-1}$ = 5i.**Example 2:** Evaluate i^{99}.**Solution:** i^{99}

= i^{(96+3)}

= i^{(4*24)}.i^{3}

= 1* (-i) [i^{4n} = 1; i^{3} = -i]

= – i **Example 3:** If z_{1} = 2 + 5i and z_{2} = 3 – 2i, find z_{1}× z_{2}.**Solution:** z_{1}× z_{2} = (2 + 5i) (3 – 2i)

= 6 + 15i – 4i – 10i^{2}

= 6 + 11i + 10

= 16 + 11i

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