Simplify (2 – 14i)(2 + 14i)

A complex number is a term that can be shown as the sum of real and imaginary numbers. 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 the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z) in form of a complex number.  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,

• 6 + 2i is a complex number, where 6 is a real number (Re) and 2i is an imaginary number (Im).
• 9 + 3i is a complex number where 9 is a real number (Re) and  3i is an imaginary number (Im)

Algebraic Operations on Complex number 

A real number and imaginary number combination are called a Complex number. There are four types of algebraic operation of complex numbers,

• Addition of Complex Numbers   

In this operation, we know that a complex number is of the form z = p + iq where a and b are real numbers. Now, consider two complex numbers z₁ = p₁ + iq₁ and z₂ = p₂ + iq₂. Therefore,  the addition of the complex numbers z₁ and z₂.

z₁ + z₂ = (p₁ + p₂) + i(q₁ + q₂)

Some more identities are: 

1. z₁ + z₂ = z
2. z₁ + z₂ = z₂ + z₁
3. (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
4. z + (-z) = 0
5. (p + iq) + (0 + i0) = p + iq

The resulting complex number real part is the sum of the real part of each complex number. The resulting complex number imaginary part is equal to the sum of the imaginary part of each complex number.

• Subtraction of Complex Numbers

In this operation of the complex numbers z₁ = p₁ + iq₁ and z₂ = p₂ + ib₂, therefore the difference of z₁ and z₂  which is z₁-z₂ is defined as,

z₁ – z₂ = (p₁ – p₂) + i(q₁ – q₂)

• Multiplication of Complex Numbers 

In this operation of multiplication of Two Complex Numbers. We know that (x + y)(z + w).

= xz + xw + zy + zw

Similarly, the complex numbers z₁ = p₁ + iq₁ and z₂ = p₂ + iq₂

To find z₁z₂:  

z₁ z₂ = (p₁ + iq₁)(p₂ + iq₂)

z₁ z₂ = p₁ p₂ + p₁ q₂i + q₁ p₂i + q₁q₂i²

As we know,  i² = -1,  

Therefore,

z₁ z₂ = (p₁ p₂ – q₁ q₂ ) + i(p₁ q₂ + p₂ q₁ )

Some identities are: 

1. z₁ × z₂ = z
2. z₁.z₂ = z₂.z₁
3. z₁(z₂.z₃) = (z₁.z₂)z₃
4. z₁(z₂ + z₃) = z₁.z₂ + z₁.z₃

• Division of Complex Numbers  

In this operation of complex number z₁ = p₁ + iq₁ and z₂ = p₂ + iq₂, therefore, to find z₁/z₂, we have to multiply the numerator and denominator with the conjugate of z₂.

The division of complex numbers:

Let z₁ = p₁ + iq₁ and z₂ = p₂ + iq₂,  

z₁/z₂ = (p₁ + iq₁)/(p₂ + iq₂)

Hence, (p₁ + iq₁)/(p₂ + iq₂) = [(p₁ + iq₁)(p₂ – iq₂)] / [(p₂ + iq₂)(p₂ – iq₂)]

(p₁ + iq₁)/(p₂ + iq₂) = [(p₁p₂) – (p₁q₂i) + (p₂q₁i) + q₁q₂)] / [(p₂² + q₂²)]

(p₁ + iq₁)/(p₂ + iq₂) = [(p₁p₂) + (q₁q₂) + i(p₂q₁ – p₁q₂)] / (p₂² + q₂²)

z₁/z₂ = (p₁p₂) + (q₁q₂) / (p₂² + q₂²) + i(p₂q₁ – p₁q₂) / (p₂² + q₂²)

Simplify (2 – 14i)(2 + 14i)

Solution:

Given: (2 – 14i)(2 + 14i)  

= {4 + 28i – 28i – 196i²}

= (4 +196)

= 200 + 0i

Similar problems

Question 1: Solve (1 – 5i) / (-3i)?

Solution: 

Given: (1 – 5i) / (-3i)

Denominator standard form i.e -3i = 0 – 3i

Conjugate of denominator 0 – 3i = 0 + 3i

Multiply with the conjugate,

Therefore, {(1 – 5i) / (0 – 3i)} × {(0 + 3i)/(0 + 3i)}

= {(1 – 5i)(0 + 3i)} / {0 – (3i)²}

= {3i – 15i²} / {0 – (9(-1))}

= {3i – 15 (-1)} / 9

= (3i +15) / 9        

= 5/3 + (1/9)i

Question 2: Perform the indicated operation and write the answer in standard form: (a + bi) (c + di).

Solution:

Given: (a + bi) (c + di)                 

= ac + bci + adi+ bdi² 

= (ac – bd) + i(bc + ad)

Question 3: Perform the indicated operation and write the answer in standard form: (4 + 4i) × (3 – 4i).

Solution:

(4 + 4i) × (3 – 4i)

= (12+ 12i – 16i – 16i2 )

= 12 – 4i +16

= 28 – 4i

Question 4: Perform the indicated operation and write the answer in standard form: (5 + 4i) × (6 – 4i).

Solution :

= (5 + 4i) × (6 – 4i)

= (30 – 20i + 24i – 16i²)

= 30 + 4i + 16

= 46 + 4i

Question 5: What is the answer to the following problem, (-5i)(4i)(-2).

Solution: 

Given: (-5i)(4i)(-2)

= -5i × 4i × (-2)

=  -20i² × -2 {i² = -1}

= -20 (-1) × -2

= 20 × -2

= -40 + 0i

Question 6: If z₁, z₂ are (1 – i), (-2 + 4i) respectively, find Im(z₁z₂/z₁).

Solution: 

Given: z₁ = (1 – i)

z₂ = (-2 + 4i)

Now to find Im(z₁z₂/z₁),

Put values of z₁ and z₂

Im(z₁z₂/z₁) = {(1 – i) (-2 + 4i)} / (1 – i)

= {(-2 + 4) + i(2 + 4)} / (1 + i)

= {(2 + 6i) /(1 + i)}                          

= {(2 + 6i) (1 + i)} / {(1 + i)(1 – i)}

= {(2 + 6) + i(6 – 2)} / (1 + 1)

= 4 + 2i

Therefore, Im(z₁z₂/z₁) = 2