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# Additive Identity and Inverse of Complex Numbers

• Last Updated : 11 Mar, 2022

The combination of a real number and an imaginary number is termed a Complex number. These are the numbers that can be written in the form of a + ib. where a and b both are real numbers. It is denoted by z. Here in complex number form the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z).  It is also called an imaginary number. In complex number form a + bi ‘i’ is an imaginary number called “iota”. The value of i is (√-1) or we can write as i2 = -1.

For example,

• 3 + 4i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
• 2 + 5i is a complex number where 2 is a real number (Re) and  5i is an imaginary number (im)

Rules of imaginary numbers

1. i = √-1
2. i2 = -1
3. i3 = -i
4. i4 = 1
5. i4n = 1
6. i4n – 1 = -1

### Additive identity and inverse of complex numbers

Solution:

Additive identity property states that when a number is added to zero will give the same number in result .

Here , “Zero” is called the identity element, (also known as additive identity) If we add any complex number with zero, the resulting number will be the same complex number. This is true for any real numbers, complex numbers and even for imaginary numbers.

Suppose, ‘a’ is any real number, then

x + 0 = x =  0 + x

For Example: Let assume z = a+ib , then as per the property z + 0= z

Therefore, (a + ib ) + (0 + 0i)

= a + bi + 0 + 0i

=  a + bi

= z, for all z ∈ C.

Hence, The additive identity of complex numbers is written as (x + yi) + (0 + 0i) = x + yi.

So, the additive identity is 0 + 0i.

An additive inverse of a complex  number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.

Suppose, x is the original number, then the additive inverse of x will be minus of x i.e., -x , such that;

x + (-x) = x – x = 0

It is also called the opposite of the number, negation of number or changed sign of original number. Suppose the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero.

Therefore, the additive inverse of A + iB is  -(A + iB)

Example: Additive inverse of 3 + 3i is -(3 + 3i)

3 + 3i + [-(3 + 3i)]

= 3 + 3i – 3 – 3i

=  0

Hence proved the property .

### Sample Problems

Question 1: Find the additive inverse of complex number 5 + 5i?

Solution:

An additive inverse of a complex number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.

So here the additive inverse of complex number 5 + 5i is -(5 + 5i)

= -5 – 5i

Question 2: Find the additive inverse of 8+3i and prove the identity?

Solution:

An additive inverse of a complex  number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.

So here the additive inverse of complex number 8 + 3i is -(8 + 3i)

= -8 – 3i

Now to prove identity the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero .

Therefore, 8 + 3i + (-8 – 3i)

= 8 + 3i – 8 – 3i

= 0

Hence proved

Question 3: Prove the additive identity complex number 7 + 4i?

Solution:

Additive identity property states that when a number is added to zero will give the same number in result. The additive identity of complex numbers is written as (x + yi) + (0 + 0i) = x + yi.

So now, to prove additive identity for complex number 7 + 4i

= (7 + 4i) + (0 + 0i)

= 7 + 4i

Hence proved

Question 4: Simplify {(-3 – 5i) / (2 +2i)}?

Solution:

Given {(-3 – 5i) / (2 + 2i)}

Conjugate of denominator 2 + 2i is 2 – 2i

Multiply the numerator and denominator with the conjugate,

Therefore, {(-3 – 5i) / (2 + 2i) } × {(2 – 2i) / (2 – 2i)}

= {-6 – 6i – 10i +10i2} / {22 – (2i)2} {Difference of squares formula. i.e; (a+b)(a-b) = a2 – b2}

= {-6 – 6i – 10i + 10(-1)} / {4 – 4(-1)} {i2 = -1}

= {-6 – 6i – 10i – 10} / {4 + 4}

= (-16 – 16i) / 8

= -16 /8 – 16i /8

= -2  – 2i

Question 5:  Find the additive inverse of complex number 3 + 5i and prove it?

Solution:

An additive inverse of a complex number is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.

So here the additive inverse of complex number 3 + 5i is -(3 + 5i)

= -3 -5i

To prove: the additive inverse of A + iB should be a value that on adding it with a given complex number will give a result as zero .

Therefore: 3 + 5i + (-3 – 5i)

= 3 + 5i – 3 + 3i

= 0

Hence proved

Question 6: Simplify (3 + 4i) / (3 + 2i) and Find its additive inverse?

Solution:

Multiplying the numerator and denominator with the conjugate of denominators.

= ((3 + 4i) × (3 – 2i)) / ((3 + 2i) × (3 – 2i))

= (9 – 6i + 12i – 8i2) / {9 – (2i)2}

= (17 + 6i) / (13)

= (17 + 6i) / 13

= 17/13 + 6i/13

Now Additive inverse of complex number,

17/13 + 6i/13 is – (17/13 + 6i/13)

= -17/13 – 6i/13

Question 7: Perform the following operation and find the result in form of a + ib and find its inverse?

• (2 – √-25) / (1 – √-16)

Solution:

Given: (2 – √-25 ) / (1 – √-16)

(2 – √-25) / (1 – √-16) = {2 – (i)(5)} / {1 – (i)(4)}, {i = √-1}

= (2 – 5i ) / (1 – 4i)

= {(2 – 5i) / (1 – 4i)} × {(1 + 4i) / (1 + 4i)}

= {(2 – 5i) (1 + 4i)} / {(1 – 4i) (1 + 4i)}

= {2 + 8i – 5i – (20i2)} / {(1 – 16i2)} {i2 = -1}

= {2 + 3i + 20} / {1 – 16(-1)}

= (22 + 3i) / (1 + 16)

= (22 + 3i)/17

= {(22/17) + (3i/17)}

= 22/17 + 3i/17

Now Additive inverse of complex number of 22/17 + 3i/17

= -(22/17 + 3i/17)

= -22/17 – 3i/17

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