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CBSE NOTES ⇒ Class 11th ⇒ Mathematics ⇒ 5. Complex Number and Quadratic Equations

## 5. Complex Number and Quadratic Equations

### 7. Algebra of Complex Numbers

(A)   Addition of Complex numbers

Let z1 = a + ib and z2 = c + id be any two complex numbers. Then, the sum z1 + z2 is defined as follows:

z1 + z2 = (a+ib)+(c+id) = (a + c) + i (b + d), which is again a complex number.

Ex: z1 =2 + i3 and z2 = 7 + i5

z1+ z2 = (2+7) +i(3+5)

= 9+i8

Alternative Method:

z1+ z2 = 2+i3 +7+i5

= 2+7+i3+i5

= 9+i(3+5)

= 9+i8

(B) Subtraction of Complex Numbers

Given any two complex numbers z1 and z2, the difference z1z2 is defined as follows:

z1z2 = (a+ib) – (c+id) = (a–c)+i(b–d)

For example: z1= 4+i3,  z2 =3 + i7

z1z2 =(4–3)+i(3–7)= 1–i4

Note: {4 and 3are like term and i3 and i7 are another like term}

Alternative Method:

z1z2 =(4+i3) – (3+i7)

=4+i3 – 3 – i7

=4– 3+i3 – i7

=1+i(3 – 7)

=1-i4

(C) Multiplication of Complex numbers:

z1 × z2 = (a +ib) (c+id)

=a(c+id) +ib(c+id)

= ac + iad + ibc +(bd)

= ac bd +i(ad+bc)

For example: z­1= 2+3i, z2=5 +2i

1× z2 = (2+3i)( 5 +2i)

= (2×5 – 3×2)+ i(2×2+3×5)

= 10 – 6 + i(4+15)

= 4 + 19i

Alternative Method:

1× z2 = (2+3i)( 5 +2i)

=2( 5 +2i) +3i( 5 +2i)

= 10 + 4i + 15i + 6i2

= 10 + 4i + 15i + 6(–1)

= 10 – 6 + 19i

= 4 + 19i

(D)Some Important identities:

1. (z1 + z2)2 = z12 + 2 z1 z2 + z22
2. (z1 – z2)2 = z12 – 2 z1 z2 + z22
3. (z1 + z2)3 = z13 + 3 z12 z2 + 3 z1z22 + z23
4. (z1 – z2)3 = z13 – 3 z12 z2 + 3 z1z22 – z23
5. z12 – z22 = (z1 + z2)( z1 – z2)

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