# What is the Multiplicative Identity and Multiplicative Inverse of the complex number?

• Last Updated : 11 Mar, 2022

Complex number is the sum of a real number and an imaginary number. These numbers 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). In complex number form a +bi ‘i’ is an imaginary number called “iota”.

Real numbers

The numbers represent a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are called real numbers. These are represented as Re().

Example: 14, -45, 0, 1/7, 2.8, √5, etc., are all real numbers.

Imaginary Numbers

The numbers which are not real are termed as Imaginary numbers. After squaring an imaginary number, it gives a result in negative. Imaginary numbers are represented as Im().

Example: √-5, √-7, √-11 are all imaginary numbers. here ‘i’ is an imaginary number called “iota”.

### What is the Multiplicative Identity and Multiplicative Inverse of the complex number?

Solution:

Multiplicative identity of complex number says that when the complex number is multiplied by the number 1 it will give that number as product.

“1” is the multiplicative identity of a number.  If the complex number being multiplied is 1 itself.

The multiplicative identity property of complex number is represented as: z.1 = z = 1.z

For example: Let assume z = a+ib, then as per the property z.1 = z

therefore, a+ib.1+0i

= a + 0i +bi + 0i

= a +bi

= z . for all z ∈ C.

Multiplicative inverse of any number N is represented by 1/N or N-1. It is also called reciprocal of number. The reciprocal of a number is that when it is multiplied with the original number the value equals to identity 1, or you can say that it is a method of dividing a number by its own to generate identity 1, such as N/N = 1.

When a number is multiplied by its own multiplicative inverse the resultant value is equal to 1. The multiplicative inverse of a complex number z is simply 1/z. It is denoted as:

1/z or z-1 (Inverse of z)

It is also called as the reciprocal of a complex number and 1 is called the multiplicative identity..

For example: Let assume z = a+ib, then as per the property z.1 = z, its inverse is z = 1/z

multiplicative inverse of z = a +ib = 1/a+ib

= 1/(a+ib) x (a-ib)/(a-ib)

= (a-ib ) / (a2 -b2 i2 )

= (a-ib) / (a2 + b2)

= a / (a2+b2) – {b /(a2+b2)} i

### Sample Questions

Question 1: Find the multiplicative inverse of -3 + 8i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = -3 +8i

Therefore z = 1/z

= 1 / (-3 +8i)

Now rationalizing

= 1/(-3+8i) x (-3-8i)/(-3-8i)

= (-3-8i) / {(-3)2 – 82i2}

= (-3-8i) / {9 +64}

= (-3-8i)/ (73)

= -3/73 – 8i/73

Question 2: Find the multiplicative inverse of 2 – 3i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = 2 – 3i

Therefore z = 1/z

= 1 / (2 – 3i)

Now rationalizing

= 1/(2 – 3i) x (2 + 3i)/(2 +3i)

= (2 + 3i) / {(2)2 – 32i2}

= (2 + 3i) / { 4 + 9}

= (2 + 3i)/ (13)

= 2/13 + 3i/13

Question 3: Find the multiplicative inverse of √5+3i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as: 1 / z  or  z-1 (Inverse of z)

here z = √5+3i

Therefore z = 1/z

=  1 / (√5+3i )

now rationalizing

= 1/(√5+3i ) x (√5 -3i)/(√5-3i)

= (√5+3i ) / {(√5)2 – (3)2 (i)2 }

=  (√5 +3i) / { (5) + 9}

=  (√5 + 3i)/ (14)

= √5/14 + 3i/14

Question 4: Find the multiplicative inverse of 4 – 3i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = 4 – 3i

therefore z = 1/z

=  1 / (4 – 3i)

Now rationalizing

= 1/(4 – 3i) x (4 + 3i)/(4 +3i)

= (4 + 3i ) / {(4)2 – 32i2}

= (4 + 3i) / {16 + 9}

= (4 + 3i)/ (25)

= 4/25 + 3i/25

Question 5: Find the multiplicative inverse of (5-7i)?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = 5 – 7i

therefore z = 1/z

= 1 / (5 – 7i)

Now rationalizing

= 1/(5 – 7i) x (5 + 7i)/(5 +7i)

= (5 + 7i ) / {(5)2 – 72i2}

= (5 + 7i) / { 25 + 49}

= (5 + 7i)/ (74)

= 5/74 + 7i/74

Question 6: Simplify (2-4i)(5-7i) And Find its multiplicative inverse?

Solution:

Given:  (2-4i)(5-7i)

= 10 -14i -20i +28i2

= 10 -14i -20i + 28(-1)2

= 10 – 14i – 20i +28

= 18 – 34i

Now, multiplicative inverse of 18 – 34i is

It is denoted as : 1 / z  or  z-1 (Inverse of z)

Here z = 18 – 34i

Therefore z = 1/z

=  1 / (18 – 34i)

Now rationalizing

= 1/(18 – 34i) x (18 + 34i)/(18 + 34i)

= (18 + 34i) / {(18)2 – 342i2}

= (18 + 34i) / {324 + 1156}

= (18 + 34i)/ (1480)

= 18/1480 + 34i/1480

= 9/740 + 17i/740

Question 7: Find the multiplicative inverse of (4 + 2i)?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = 4 + 2i

Therefore z = 1/z

=  1 / (4 + 2i)

Now rationalizing

= 1/(4 + 2i) x (4 – 2i)/(4 – 2i)

= (4 – 2i) / {(4)2 – 22i2}

= (4 – 2i) / {16 + 4}

= (4 – 2i)/ (25)

= 4/25 – 2i/25

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