Chapter 5

### Complex Numbers

In earlier classes, we have studied second degree equations. We have seen that if the b 2 -4ac is less than 0, then the second degree equation has no solution. What is the reason? Negative real numbers has no real root. So we need to extend the real number system to a larger number system so that negative real number has also a square root. In fact, the main objective is to solve a second degree equation in the case when the discriminant is a negative real number which is not possible in the system of real numbers. For this we have to study a new type of numbers called imaginary numbers.

Let us start…….

**Welcome all**

### Imaginary number

An imaginary number is the square root of a negative real number. It can be written as a real number multiplied by the imaginary unit i(iota)

√(-1) = i

i2 =-1 i3 = -I i4 = 1

i0= i4 = i8 = i12 =…= 1 i4n = 1

√(-9) =√(9×-1) = √9 ×√(-1 ) = 3i

√(-4 ) = 2i

√(-5) = i√5

in can be written as ik, where k is the remainder obtained by dividing n by 4

i10 = i2 = -1 (2 is the remainder when 10 is divided by 2)

i51 = i3= -i

i-39 = 1/i39 = 1/i3 = i4/i3 = i

i9 + i19 = i1 + i3 = i-i = 0

Prove that i^n + i^(n+1) + i^(n+2) + i^(n+3) = 0

i^n + i^(n+1) + i^(n+2) + i^(n+3) am+n = am × an

= in + in×i1 + in×i2 + in×i3

= in(1+i1+i2+i3)

= in(1+i-1-i) = 0

### Complex Number

A number of the form a+ib, where a and b are real numbers is called a complex number. Complex numbers are generally denoted by z A is called real part(Rez) and b is called imaginary part(Imz)

Eg : z = -2 +√3 i Rez = -2, Imz = √3

Every real number is a complex number

7 = 7 + 0i

Set of all complex numbers is denoted by C. R is a subset of C

A complex number with real part 0 is called purely imaginary complex number

Eg: i,3i,-i,-2i

If two complex numbers are equal, their real parts are equal and imaginary parts are equal

Q. If 4x+i(3x-y) = 8-6i, find x and y

Equating the real part

4x = 8 x = 8/4 = 2

Equating the imaginary part

3x-y = -6

3×2 – y = -6

6 – y = -6

Y = 6 + 6 = 12

### Conjugate of a complex number

Conjugate of z = a+ib is denoted by z and it is defined as z = a-ib

Eg: z = -1-i, z = -1+i

### Modulus of a complex number

Modulus of z = a+ib is denoted by |z| and is defined as

|z| = √(a^2+b^2 )

Eg : z = 3-4i

|z|= √(3^2+〖-4〗^2 )

= √(9+16 )= 5

Q. Find the sum i + i2 + i3 + …. + i100

Ans:

Sequence is a G.P with a = i, r = i

Sn = (a(r^n-1))/(r-1)

S100 = (i(i^100-1))/(i-1)

= (i(1-1))/(i-1) = 0

Q. √(-25 )+ 3√(-4 ) + 2√(-9 )

Ans:

= 5i + 3×2i + 2×3i

= 5i + 6i + 6i

= 17i

### Operations

(2+3i) + (-1+2i) = (2+ -1) + (3i+2i) = 1 + 5i

(1-i) – (4-2i) = 1 –i -4+2i = -3-i

(2+i)2 = 22 + 2.2.i + i2 = 4 + 4i -1 = 3 + 4i

(2+3i)(3-2i) = 6-4i+9i-6i2 = 6+5i-6×-1 = 6+5i+6= 12 + 5i

Negative real numbers have no real roots

Q. write (2+3i)/(4+5i) in a+ib form

Ans:

((2+3i)(4-5i))/((4+5i)(4-5i))

= (8-10i+12i+15)/(〖(4)〗^2- 〖(5i)〗^2 )

= (23+2i)/(16+25)

= (23+2i)/41

= 23/41 + (2 i)/41

Q. (1+i)4

(1+i)2 = 12 + 2×1×i + i2

= 1+2i-1= 2i

(1+i)4 =〖〖[(1+i)〗^2]〗^2

= 〖(2i)〗^2

= 22×i2

= 4×-1

= -4

= -4+0i

Q. (3-4i)/((4-2i)(1+i))

= (3-4i)/(4+4i-2i+2

= (3-4i)/(6+2i)

= ((3-4i)(6-2i))/((6+2i)(6-2i))

= (18-6i-24i-8)/(6^2- 〖(2i)〗^2 )

= (10-30i)/(36+4)

= (10-30i)/40

= 10/40 – 30i/40 = 1/4 – 3i/4

Q. Find the least positive value of n such that ((1+i)/(1-i))n = 1

(1+i)/(1-i)

= ((1+i)(1+i))/((1-i)(1+i))

= (1+i+i+ i^2)/(1^2-i^2 )

=(1+2i-1)/(1+1)

= 2i/2

= i

((1+i)/(1-i))n = 1

in = 1

i4 = 1

n = 4

Q. Find the multiplicative inverse of 4 + 3i

Multiplicative inverse of a complex number is its reciprocal

1/(4+3i) = ((4-3i))/((4+3i)(4-3i)) = ((4-3i))/(4^2-〖(3i)〗^2 ) = (4-3i)/(16+9) = (4-3i)/25 = 4/25 – 3i/25

### Geometrical representation of a complex number

A complex number z = a+ib can be represented as a point (a,b) in a plane called Complex plane or Argand plane. X axis is known as real axis and y axis is known as imaginary axis. A real number is represented on real axis and purely imaginary complex number is represented on imaginary axis If Z is in first quadrant, its conjugate is in fourth quadrant. If Z is in second quadrant, its conjugate is in third quadrant. The modulus of a complex number is the distance from the origin.

(+,+) I

(-,+) II

(-,-) III

(+,-) IV

###

Polar form of a complex number

Every nonzero complex number can be represented in the form r(cosϴ+isinϴ ). This form is called polar form of a complex number, where r is the modulus and ϴ is the argument of the complex number.

a+ib = r(cosϴ+isinϴ)

r= √(a^2+b^2 )

tanα = |b/a|

in first quadrant ϴ = α, in second quadrantϴ = π –α, in third quadrant ϴ = α – π, in fourth quadrant ϴ = – α

Square root of a complex number z = a + ib

±[√((|Z|+a)/2) + i√((|Z|-a)/2) ]

If imaginary part of given complex number is –ve, take (+,-)and (-,+)

If imaginary part of given complex number is +ve, take (+,+)and (-.-)

Q. Find the square root of -7-24i

Ans:

a = -7, b= -24

|z|= √(a^2+b^2 )

= √(〖(-7)〗^2+〖(-24)〗^2 )

= √(49+526)

= √625 = 25

±[√((|Z|+a)/2) + i√((|Z|-a)/2) ]

±[√((25-7)/2) + i√((25+7)/2) ]

±[√(18/2) + i√(32/2) ]

±[√9 + i√16 ]

±[3 + 4i ] 3 + 4i, -3 – 4i,3 – 4i,-3 + 4i

Square roots are 3-4i and -3+4i

**Q. Solve the quadratic equation √5 x2 + x + √5 = 0**

Ans

a = √5 , b = 1, c = √5

b2 – 4ac = 12 – 4×√5×√5

= 1 – 20 = -19

X = (-b±√(b^2-4ac))/2a

= (-1±√(-19))/(2√5)

= (-1±i√19)/(2√5)

Exercise

1.Find i49 + i68 + i89 + i110

2. Find √(-4) ×√(-9)

3. Write (1-i)4 in a+ib form

4.

5. Write the a+ib form of (3-√(-16))/(1-√(-9))

6. If x+iy = (a+ib)/(a-ib). Prove that x2+y2=1

Hint. |x+iy| = (|a+ib|)/(|a-ib|)

7. Find the square root of 5 + 12i

8. Write the polar form of ((2+i))/((1+i)(1-2i))

9. Solve ix2-x+12i = 0

10. Write the a+ib form of (5- i√(3 ))/(4+2√(3 ) i)

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