__Algebraic Operations on Complex Numbers__**
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In performing operations with complex numbers we can proceed as in the algebra of real numbers replacing i^{2} by -1 when it occurs.

**1. Addition of Complex Numbers
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*(a+bi) + (c+di) = (a+c) + (b+d)i
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**Example:**

*z1 = 3 - 4i and z2 = -2 + 2i*

z1 + z2 = (3 + (-2)) + (-4 + 2)i = 1 -2i*
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**2. Subtraction of Complex Numbers
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*(a+bi) - (c+di) = (a-c)+ (b-d)i
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**Example:**

*z1 = 3 - 4i and z2 = -2 + 2i*

z1 - z2 = (3 - (-2)) - (-4 - 2)i = 5 +6i**
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**3. Multiplication of Complex Numbers
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*(a+bi)(c+di) = ac+bci+adi+bdi*^{2}

* = (ac-bd)+(ad+bc)i
*

*
*

**Example:**

*z1 = 3 - 4i and z2 = -2 + 2i*

z1 ∙ z2 = (3 ∙ (-2)) + (-4 ∙ (-2))i + (3 ∙ 2)i + (-4 ∙ 2)i^{2} = -6 + 14i -8i^{2}*
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**4. Division of Complex Numbers
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*(a+bi)/(c+di)=(a+bi)/(c+di)×(c-di)/(c-di) *

=(ac-adi+bci-bdi^{2})/(c^{2}-d^{2}i^{2})

=(ac+bd+(bc-ad)i)/(c^{2}-d^{2}i^{2})

=(ac+bd)/(c^{2}+d^{2} )+(bc-ad)/(c^{2}+d^{2}) i

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**Example:**

*z1 = 3 - 4i and z2 = -2 + 2i*

z1 / z2 = (3 ∙ (-2) + (-4 ∙ 2))/((-2)^{2} + 2^{2}) + (-4 ∙ (-2) - 3 ∙ 2)/((-2)^{2} + 2^{2})i = -14/8 + (2/8)i = -7/4 + i/4

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Inequalities in imaginary numbers are not defined. As for example *z >0, 4 + zi < 2+4i* are meaningless in complex numbers.

In real numbers, if *a*^{2} + b^{2}=0 then *a=b=0*; however in complex numbers,

*z*_{1}^{2} + z_{2}^{2} does not imply *z*_{1} = z_{2} = 0

Complex conjugation: The complex conjugate of *a + bi is a − bi*. The importance of the complex conjugate lies in the fact that the product of a complex number and its conjugate is a real number.

*(a + bi)(a − bi) = (a + b**)
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Additive identity: Additive identity is the complex number that represents the notion of zero for addition of complex numbers is *0 + 0i*.

(a + bi) + (0 + 0i) = a + bi

Multiplicative identity: Multiplicative identity is the complex number that represents the notion of one for multiplication of complex numbers is 1+ 0i.

(a + bi)(1+ 0i) = a + bi

**Example 1**: Simplify *(2**√(-9) -3 )(3√(-16) - 1)*

*(2**√(-9) -3 )(3√(-16) - 1)
*

*=(2 · 3i - 3)(3 ·**4**i - 1)
*

*=(6i - 3)(12i - 1)
*

*= 72i*^{2} – 36i – 6i + 3

*= -69 – 42i
*

**Example 2**: Express in the form *a + bj* given

**Example 3**: Perform the division for the complex numbers

__Solution__: The given complex numbers are

We have to reduce the denominator as well as the numerator by multiplying the numerator and denominator both by the complex number 10+2i.

Use the formula *(a+ b) (a-b) = a*^{2}- b^{2} to reduce the denominators. Use the formula *(a+ b) *^{2}= a^{2}+2ab+ b^{2} to reduce the numerators.

Add and also square the terms to get the values.

In the complex numbers the value for i^{2} is -1.

Substitute the value of i^{2} in the above equation to reduce the denominators.

**Example 4**: Find the product of the complex numbers 1+ *i* and √ 3- *i* in polar form.

__Solution__: We can easily find that the polar form of 1+ *i* and √ 3- *i* are respectively,

We know that *z*_{1}z_{2} = r_{1}r_{2}(cos*(θ*_{1}+θ_{2}) + isin*(θ*_{1} + θ_{2})). This formula says that to multiply two complex numbers we need to multiply the moduli and add the arguments. Therefore multiplying 1+ *i* and √ 3- *i* using the above rule, we will get:

**Example 5**: Simplify *i*^{64,002}.

__Solution__: *i*^{64,002} =* i*^{64,000 + 2} = *i*^{4 · 16,000 + 2} = *i*^{2} = –1.