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# Simplifying Complex Numbers

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. i2 = -1
3. i3 = -i
4. i2n = 1;   where n is any natural number.

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 + bdi2 = ac + bci + adi - bd = (ac - bd) + i(bc + ad)

Example: Solve (3 + 2i) (5 – i)
Solution: (3 + 2i) (5 – i)
= 15 + 10i – 3i – 2i2
= 15 + 7i – 2(-1)        [i2 = -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) = a2 – (bi)2 = a2 – b2i2 = a2 + b2
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.
Example: $\frac{2+3i}{4-5i}$

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 i99.
Solution: i99
= i(96+3)
= i(4*24).i3
= 1* (-i)        [i4n = 1;  i3 = -i]
= – i

Example 3: If z1 = 2 + 5i and z2 = 3 – 2i, find z1× z2.
Solution: z1× z2 = (2 + 5i) (3 – 2i)
= 6 + 15i – 4i – 10i2
= 6 + 11i + 10
= 16 + 11i

 Related Calculators Simplify Complex Numbers Calculator Simplify Complex Fractions Calculator Adding Complex Numbers Calculator Dividing Complex Numbers Calculator

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