Skip to content
Related Articles

Related Articles

Simplify (2 – 14i)(2 + 14i)

View Discussion
Improve Article
Save Article
Like Article
  • Last Updated : 15 Mar, 2022

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+ bdi 

= (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

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!