CBSE Class 11 Maths Revision Notes Chapter 5

Class 11 Mathematics Revision Notes for Chapter-5 Complex Numbers and Quadratic Equations 

Class 11 Chapter 5 Mathematics Notes provided by Extramarks will help students in grasping concepts quickly and efficiently. The notes are explained step by step for all NCERT chapters, thereby ensuring that the students gain conceptual clarity and clear their exams with flying colours.

Chapter 5 Mathematics Class 11 Notes for Complex Numbers and Quadratic Equations contain important points for different mathematical formulae and theorems. Extramarks provides notes for all NCERT chapters explaining all the topics with proper explanations. This will help students with the revisions and preparations before examinations.

Topics and Subtopics in Class 11 Mathematics Notes of Complex Numbers and Quadratic Equations 

Definition 

When a given number is in the form of a+ib, where a, b∈R and i=√−1 it is known as a complex number and such number is generally denoted by ‘z’. 

z= a+ib 

Where, 

a= real part of the complex number and, 

b= imaginary part of the complex number.

1.1 Conjugate of a Complex Number 

The conjugate of a complex number, z̄ = a + ib,

is represented by the symbol ‘z̄’.

Whose value is specified as z = a − ib.

  1. Algebra of Complex Numbers 

Addition:

z1 + z2 = (a + bi) + (c + di)

= (a + c) + (b + d)i

Subtraction:

z1 − z2 = (a + bi) − (c + di)

= (a − c) + (b − d)

Multiplication:

z1⋅ z2 = (a + bi)(c + di)

= a(c + di) + bi(c + di)

= ac + adi + bci + bdi2

= ac − bd + (ad + bc)i (∵ i2 = −1) 

Note…

  1. a + ib = c + id

⇔ a = c&b = d

  1. i4k + r =  1; r = 0

                     i; r = 1

                  −1; r = 2

                   −i; r = 3

  1. √b√a = √ba is only possible if at least one

of either a or b is non-negative.

  1. Argand Plane 

Any complex number with the formula z = a+ib can be represented by a single point P(a, b) in the argand plane.

3.1 Modulus and Argument of Complex Number 

Think about the complex number z = a + ib.

(i) The modulus of the complex number z is the distance of the complex number z from the origin.

It is shown by the equation r = |z| = (a2 + b2)

(ii) This time, i.e. here, θ i.e. the angle made by ray OP with positive direction of the real axis is called the argument of z.

Note.

z1>z2 or z1<z2 has no relevance yet |z1|>|z2| or |z1|<|z2| holds relevance.

3.2 Principal Argument 

The main argument ′θ′ of complex numbers z=a+ib is referred as the principal argument of zif −π<θ⩽π. Take into account, that tan = b/a, and θ be the arg(z). The principal argument is provided by −π+α and −α respectively.

  1. Polar Form 

a=rcosθ & b=rsinθ where r=|z| and θ=arg(z)∴z=a+ib=r(cosθ+isinθ)

Note…

Another way to express a complex number z, also known as Euler’s form is z=re

Where,

θ=arg(Z)&r=|Z|

  1. Some Important Properties

Some of them are: 

  1. z+z̄= 2Re(z)
  2. z−z¯= 2iIm(z)
  3. |z|=0⇒z=0 
  4. z.z̄=|z|2
  5. |z̄|=|z|=|−z|

 

  1. De-Moivre’s Theorem

Statement: cos nθ + is in nθ is the value or one of the values of (cosθ+isinθ)n. Here, ‘n’ is a rational number or an integer. The theorem is very helpful in finding out the roots of any complex quantity.

  1. Cube Root of Unity 

Roots of the equation x3=1 are known as cube roots of unity.

<Equations>

  1. ‘n’ nth Roots of Unity

<Equations>

Note. 

  1. It is possible to obtain the ‘n’ nth roots of unity by taking any n consecutive integral values of k.
  2. Zero is equal to the sum of the unity’s nth roots, nN.
  3. The vertex points of a regular polygon (with n sides) encircled by a unit circle, each of which is centred at the origin, are the positions represented by the “n”th roots of unity.
  4. Square Root of Complex Number 

Suppose x + iy = a + ib

By squaring both sides, we obtain

(x+iy)2=a+ib 

i.e., x2−y2=a, 2xy=b

Solving these equations, we can find out square roots of z.

  1. LOCI in Complex Plane

(i) |z−z0|= a denotes the circumference of a circle with a radius of a and centre of zo

(ii) |z−z0|< a denotes the circle’s interior.

(iii) |z−z0|> a denotes this circle’s edge.

(iv) |z−z1|=|z−z2| represents ⊥ the bisector of a segment with endpoints z1 and z2.

(v) arg(z)=θ is a ray with a real axis that is inclined at ∠θ with respect to the origin (excluded).

  1. Vectorial Representation of a Complex

Each complex number can be considered like it is the position vector of that point. (Example)

Note.

  1. Some Important Results

(i) If z1 and z2 are two complex numbers, then the space between z1 and z2 is |z2−z1|.

(ii) This equation describes the line that connects z1 and z2:

z−z / z̄−z̄ = z−z2 / z̄−z̄2

(iii) The diameter of the circle on the line segment connecting z1 and z2 is:

(z−z1) (z̄¯−z̄2) + (z−z2) (z̄−z̄1)= 0

(iv) If complex numbers z1,bz2, z3 and z4 are represented by four points namely, A,B,C and D then:

AB||CD if z4−z3 / z2−z1 is wholly real;

AB⊥CD if z4−z3 / z2−z1 is wholly imaginary.

Important Identities 

Some of them are:

(i) x2+x+1=(x−ω)(x−ω2)

(ii) x2−x+1=(x+ω)(x+ω2)

(iii) x2−xy+y2=(x+ωy)(x+yω2)

(iv) x3−y3=(x−y)(x−yω)(x−yω2)

(v) x3+y3+z3−3xyz=(x+y+z) (x+ωy+ω2z) (x+ω2y+ωz)

  1. Quadratic Expression 

A quadratic expression in x has the usual form, f(x)= ax2+bx+c, where a,b,c∈R and a≠0. A quadratic equation in x has the general form, ax2+bx+c=0, where a, b, cR and a≠0.

  1. Roots of Quadratic Equation 

Roots of quadratic equation are the values of variables satisfying the given quadratic equation.

Note.

y = (ax2+bx+c) ≡ a(x−α)(x−β) = a[x+(b/2a)]2− D/4a

  1. Nature of Roots 

(a) Take into account the quadratic equation ax2+bx+c= 0 where a, a,bb, c∈R and a≠0 then;

(i) D>0⇔ roots are distinct and real (unequal).

(ii) D=0⇔ roots are coincident and actual (equal).

(iii) D<0⇔ roots are imaginary.

(iv) If a quadratic equation has two roots, one of them must be p+iq, then the other must be the conjugate  p−iq and vice versa. (p, q∈R and i= √-1).

(b) Think about the quadratic formula ax2+bx+c=0 where a, b, c∈Q and a≠0 then;

(i) Roots are logical and unequal, is D>0 and is a perfect square. 

(ii) If α= p+√q is one root in this situation, (where p is rational and √q is a surd) then the other root must be the conjugate of it i.e., β= p−√q and vice versa.

  1. Graph of Quadratic Equation 

Now, think about the quadratic expression, y=ax2+bx+c, a≠0 and a,b,c∈R then;

(i) In all cases, the graph between x and y is a parabola. When a>0, the parabola has a concave upward shape, and when a<0, it has a concave downward shape.

(ii) y>0∀x∈R, only if a>0 and D<0

(iii) y<0∀x∈R, only if a<0 and D<0

  1. Solution of Quadratic Inequalities 

ax2+bx+c>0(a≠0) 

(i) If D>0, then the equation ax2+bx+c=0 has two different roots (x1<x2)

Then a>0⇒x∈(−∞,x1)∪(x2,∞)

a<0⇒x∈(x1,x2)

(ii) The interval method is a simple way to solve inequalities of the form P(x) / Q(x) ≷ 0 (wavy curve).

  1. Maximum and Minimum Value of Quadratic Equation 

Maximum and minimum value of y=ax2+bx+c occurs at x=−(b/2a) according as:

For a>0, we have:

y∈ [4ac−b2 / 4a,∞)

ymin= −D/4a at x=−b/2a, and ymax→∞

For a<0, we have:

y∈ (−∞, 4ac−b2 /4a]

ymax=−D/4a at x=−b/2a, and ymin→∞

  1. Theory of Equations 

If α123,……,αn are the roots of the nth degree polynomial equation:

f(x)= a0xn+a1xn−1+a2xn−2+……+an−1x+an= 0

where a0,a1,…….an are all real and a0≠0

Then,

∑α1= −a1/a0

∑α1α2= a2/a0

∑α1α2α3= −a3/a0;

α1α2α3….αn= (−1)n an/a0

  1. Location of Roots 

Let f(x)=ax2+bx+c, where a>0 and a, b, c∈R

(i) Conditions for both the roots of f(x)=0 to be greater than a specified number ‘k’ are:

D⩾ 0 and f(k)> 0 and (−b/2a)> k

(ii) The following conditions must be met for either one of the two roots of the equation f(x)=0 to be on either side of the number k:

af(k)<0

(iii) The interval (k1,k2) must include exactly one root of f(x)=0 in order for k1<x<k2 to exist:

D>0 and f(k1).f(k2)<0

(iv) Conditions that both the roots of f(x)=0 to be confined between the numbers k1 and k2 are (k1<k2):

D⩾0 and f(k1)>0 and f(k2)>0 and k1<(−b/ 2a)<k2

  1. Maximum and Minimum Values of Rational Numbers 

Here, we will discover the values that may be obtained for real values of x by a rational expression of the form a1x2+b1x+c1/ a2x2+b2x+c2.

  1. Common Factors

(a) Only One Common Root

Let α be the common root of ax2+bx+c= 0 and a′x2+b′x+c′= 0, such that a, a′≠0 and ab′≠a′b. Then, the condition for one common root is:

(ca′−c′a)2=(ab′−a′b)(bc′−b′c)

(b) Two common roots

Considering, α, β be the two common roots of

ax2+bx+c=0 and a′x2+b′x+c′=0 such that a, a′≠0.

Then, the condition for two common roots is: a/a′= b/b′= c/c′

  1. Resolution Into Two Linear Factors 

A quadratic function f(x,y)=ax2+2hxy+by2+2gx+2fy+c may be reduced to two linear elements under the following circumstances:

abc+2fgh−af2−bg2−ch2=0 

  1. Formation of Polynomial Equations 

If the roots of the nth degree polynomial equation are α123,……,αn, then the equation is

xn−S1xn−1+S2xn−2+S3xn−3+……+(−1)nSn=0

where Sk represents the total of the roots’ products when taken k steps at a time. 

Particular Cases 

(a) Quadratic Equation: If the quadratic equation’s roots are α, β, then the equation is :

x2−S1x+S2= 0 i.e. 

x2−(α+β)x+αβ=0

(b) Cubic Equation: If the cubic equation’s roots are α, β, γ, then the equation is:

x3−S1x2+S2x−S3=0 i.e 

x3−(α+β+γ)x2+(αβ+βγ+γα)x−αβγ=0

(i) The polynomial f(x) is exactly divisible by (x) if only if α is a root of the equation f(x)=0. In other words, f(x) and (x) are factors of each other.

(ii) Every equation of nth degree (n≥1) has exactly n roots & if the equation has more than n roots, it is an identity.

  1. Transformation of Equations 

(i) By substituting 1/x for x in the above equation, an equation is created whose roots are reciprocals of the roots of the original equation.

(ii) x is changed to – x to change one equation into another whose roots are the opposite of the roots of the original equation.

(iii) x is changed to x in order to change one equation into another whose roots are the square of the roots of the original equation.

(iv) x is changed to x1/3 to change one equation into another whose roots are cubes of the roots of the original equation.

Class 11 Mathematics Notes of Complex Numbers

SECTION TOPIC
5 Complex Numbers and Quadratic Equations
5.1 Introduction
5.2 Complex numbers
5.3 Algebra of Complex Numbers
5.4 The Modulus and the Conjugate of a Complex Number
5.5 Argand Plane and Polar Representation
5.6 Quadratic Equations

What is a Complex Number?

In the case where “a” is a real number and “b” is an imaginary number, the complex number is denoted as a + ib. The requirement i2 = 1 is guaranteed by the complex number containing the letter “i.” One-dimensional number lines are referred to as the addition of such numbers. The symbol for a complex number is a + bi, and it is typically represented by the complex plane’s point (a, b).

Properties of Complex Numbers 

The properties of complex numbers state that: 

  • a and b are two real numbers where if, a + ib = 0 then b = 0, a = 0 
  • When a, b and c are the real numbers; and a + ib = c + id, then a = c and b = d. 
  • A set of three complex numbers namely z1, z2, and z3 satisfy the associative, commutative and distributive laws. 
  • When both the sum and the product of any two given complex numbers are real, the complex numbers are conjugate to each other. 
  • For two complex numbers z2 and z1 : |z1 + z2| ≤ |z1| + |z2| 
  • The sum of any two conjugate complex numbers is real. 
  • The product of any two conjugate complex numbers is real.

Importance of Chapter 5 Class 11 Mathematics Revision Notes for Complex Numbers and Quadratic Equations

Class 11 Mathematics Chapter 5 Notes provided by Extramarks follow the updated CBSE syllabus. These revision notes for Complex Numbers and Quadratic Equations have been written by subject-matter experts who have vast experience and knowledge of the field. Students can use these notes for a quick review of the chapter or note down important points for a particular chapter. 

Q.1

If α and β are different complex numbers with |β|=1,then find |βα1α¯β|.

Ans

Let α=a+ib and β=c+id   α¯=aib and |β|=c2+d2=1               c2+d2=1                                 ...1                    βα=c+idaib                            =ca+i(db)                  |βα|=(ca)2+(db)2                            =c2+a22ac+d2+b22db                            =c2+d2+a2+b22ac2db                            =c2+d2+a2+b22ac2db                            =1+a2+b22ac2db                 ...2by1                   1α¯β=1a+id¯c+id                            =1(aid)(c + id)                            =1(ac + iadibc + db)                            =(1acbd) + i(adbc)                 |1α¯β|=(1acad)2+(adbc)2                            =1+a2c2+b2d22ac+2acbd2bd+a2d2+b2c22adbc                            =1+a2c2+d2+b2c2+d22ac2bd                            =1+a2+b22ac2bd                   ...3Now          βα1αβ=1+a2+b22ac2bd1+a2+b22ac2bd=1               by 2 and 3               βα1α¯β=1

Q.2

If 1+i1im=1, then find the least positive integral value of m.

Ans

We have,1+i1i=1+i1i×1+i1+i          =(1+i)212i2          =1+2i+i21(1)=1+2i+(1)1+1          =2i2=i  1+i1im=1      im=i2         m=2Thus the least integral value of m is 2.

Q.3

Convert the complex number z=i1cosπ3+isinπ3 in the polar form.

Ans

z=i112+32i   =2i11+3i×13i13i   =2i+31+3i1+3=312+3+12iNow rcosθ=312,rsinθ=3+12On squaring and adding, we obtainr2=3122+3+122=232+14   =2×44=2r=2, which gives cosθ=3122                                 sinθ=3+122tanθ=3+131=1+1/311/3=tan45°+tan30°1tan45°tan30°=tan45°+30°   θ=π4+π6=5π12Hence the polar form is2cos5π12+i sin5π12

Q.4

Convert the complex number 3+i in the polar form and represent it in Argand Plane.

Ans

Let z=3+i=r(cosθ+isinθ)Comparing real part and imaginary part, we getrcosθ=3 and rsinθ=1On squaring and adding, we getr2cos2θ+sin2θ=32+12   r2=4    r=2  cosθ=32 and sinθ=12The value of θ satistying both the equation is by, θ=5π6Hence Z=2cos5π6+isin5π6

Q.5

Convertthecomplexnumber161+i3intopolarform.

Ans

The given complex number 161+i3=161+i3×1i31i3                                                  =16(1i3)1(i3)3=16(1i3)1+3                                                  =4(1i3)                                                  =4+i43

Let 4=rcosθ,43=rsinθBy squaring and adding, we get16+48=r2cos2θ+sin2θ    r2=64, i.e., r=8Hence cosθ=12,sinθ=32               θ=ππ3=2π3Thus, the required polar form is:8cos2π3+isin2π3

Q.6

If x+iy=a+ibaib, prove that x2+y2=1

Ans

   x+iy=a+ibaibor x+iy=(a+ib)(a+ib)(aib)(a+ib)             =a2b2+2abia2+b2             =a2b2a2+b2+2aba2+b2  xiy=a2b2a2+b22aba2+b2jNow, x2+y2=(x+iy)(xiy)                   =a2b22a2+b22+4a2b2a2+b2                   =a2+b22a2+b22=1

Q.7

Find the modulus and the argument of the following :(i)(2i)2(ii)1+i1i

Ans

i2i2=414i=34i  |34i|=9+16=5    modulus=5Since (3,4i) lies in third quadrant, therefore itsargument θ is given byθ=π+tan143ii1+i1i=1+i1i×1+i1+i=11+2i1+1=i=0+i   |0+i|=1=1    modulus =1and argument θ is given by      θ=tan110   θ=π2

Q.8

If z1=2+i,z2=23i and z3=4+5i, evaluate(i)Re(z1z2¯z3)(ii)Im(z1z2¯z3)

Ans

i z1z2¯z3=(2+i)(23i¯)4+5i              =(2+i)(2+3i)(45i)16+25              =4441+2741i  Rez1z2¯z3=4441iiImz1z2¯z3=2741

Q.9

Solve equation 2z=|z|+2i

Ans

Let z=x+iy where x,yR   |z|=x2+y2given 2z=|z|+2i  2(x+iy)=x2+y2+2i  2x=x2+y2   and 2y=24x2=x2+y2       and y=13x2=1                and y=1    x=±13          and y=1  But z=13+i does not satisty the given equation 2z=|z|+2i      z=13+i

Q.10

Solve: 5x2+x+5=0

Ans

Here the discriminant b24ac of the equation is124×5×5=120=19 the solutions are 1±1925=1±i1925

Q.11

Express in the standard form a+ib:[i18+1i25]3

Ans

i18+1i253=i4×4+2+(i)253 =i2i6×4+13 =1i3=1+i3 =1+3i+3i2+i3 =1+3i3i=2+2i =22i

Q.12 Find the multiplicative inverse of 2 + 3i.

Ans

z=2+3iz¯=23i and |z|=22+32=13 the multiplicative inverse of 2+3i is given byz1=z¯|z|2=23i13 =213313i

Q.13 If 2x + i (x – y) = 5, where x and y are real numbers, find the values of x and y.

Ans

We have 2x+ i(x – y) = 5
or 2x+ i(x – y) = 5 + 0.i
Comparing the real and imaginary parts, we get
2x = 5 and x – y = 0
⇒ x = 5/2 and x = y
Thus x = y = 5/2.

Q.14

Evaluate (1)4n+3, where nN

Ans

14n+3=i4n+3=i4ni3=1×i=i

Q.15

Show that: z+z¯=2Re(z)

Ans

Let z=a+ibThen, z¯=aibSo, z+z¯=a+ib+aib =2a=2Rez

Q.16

Express 134i in the standard form a+ib.

Ans

134i=134i×3+4i3+4i=3+4i32+42=1253+4i=325+425i

Q.17 Find the multiplicative inverse of 3 + 2i.

Ans

Let z=3+2iThen its multiplicative inverse is given by1z=13+2i=13+2i×32i32i=32i32+22=32i13=313213i

Q.18

Find the modulus of z=1+i1i.

Ans

z=1+i1i×1+i1+i =(1+i)21+1=1+i2+2i2 =11+2i2=i=0+i|z|=02+12=1

Q.19

Find the argument of 1+3i.

Ans

argz=α=tan1Im(z)Re(z) =tan131 =π3

Q.20 Solve the equation 25x2 + 9 = 0.

Ans

Here, 25x2+9=0 52x29i2=0 (5x)2(3i)2=0(5x3i)(5x+3i)=0 x=35i and x=35i

Q.21

Find the smallest positive integral value of n for which (1+i1i)n is purely a real numberwith negative real part.

Ans

We have,1+i1in=1+i1i×1+i1+in            =1+2i+i212i2n            =1+2i11+1n            =2i2n            =in  1+i1in is purely real number with -ve real partin= is purely real number with -ve real partin=1=i2    n=2

Q.22 Write the additive inverse of –3 + 4i.

Ans

The additive inverse of –3 + 4i is 3 – 4i.

Q.23 Write the conjugate of complex number –5 + 3i.

Ans

The conjugate of complex number –5 + 3i is –5 – 3i.

Q.24 Write the multiplicative inverse of i.

Ans

The multiplicative inverse of i=1i =1i×ii =ii2 =i1 =i

Q.25

Evaluate (i) 2009+(i)2010+(i)2011+(i)2012 where i=1

Ans

  2009=502×4+1    2010=502×4+2    2011=502×4+3    2012=502×4+4i 2009=i4502×i1=1×i=ii 2010=i4502×i2=1×1=1i2=1i3=ii4=1i2011=i4502×i3=1×i=ii 2012=i4502×i4=1×1=1     i2009+i2010+i2011+i2012=i1+i+1=0

Q.26

Evaluate 1+i1i2011, where i=1

Ans

We have,1+i1i2011=1+i1i×1+i1+i2011               =1+2i+i212i22011               =1+2i11+12011               =2i22011               =i2011               =i4502×i3   i4=1,i3=i               =1×i=i

Q.27

Find the smallest positive integral value of n for which (1+i1i)n is purely an imaginarynumber with negative real part.

Ans

We have,1+i1in=1+i1i×1+i1+in           =1+2i+i212i2n           =1+2i11+1n           =2i2n           =in    1+i1in is purely an imaginary number with -ve real part   in= is purely an imaginary number with -ve real part   in=i      n=3

Q.28 Find the square root of complex number 5 + 12i.

Ans

We have5+12i=94+12i          =3222+2×6i          =32+i222+2×3×2i          =32+2i2+232i          =3+2i25+12i=(3+2i)2            =±(3+2i)

Q.29

Find the value of |z| if iz3+z2z+i=0

Ans

We haveiz3+z2z+i=0On dividing both sides by i    z3+1iz21iz+1=0z3iz2+izi2=0      1i=i,i2=1z2(zi)+i(zi)=0(zi)z2+i=0(zi)=0 or z2+i=0  z=i|z|=1Also z2=i  z2=|i|  |z|2=1   |z|=1In both cases, value of |z|=1

Q.30

Put the complex number (1+i1i) in polar form.

Ans

We have,z=1+i1i=1+i1i×1+i1+i  =1+2i+i212i2  =1+2i11+1  =2i2  =i=0+i    z=0+ir=|z|=1   tanθ=yX  tanθ=10=  tanθ=tanπ2       θ=π2In polar form, z=rcos θ+i sin θ       z=1cosπ2+i sinπ2

Q.31

Calculate the value of 8+6i+86i

Ans

We have,  8+6i=91+6i           =3212+2×3i           =32+i2+2×3×i           =32+i2+23i           =3+i28+6i=(3+i)28+6i=3+iAlso,   86i = 916i           =32122×3i           =32+i22×3×i           =32+i223i           =3i286i=3i286i=3i  8+6i+86i=3+i+3i=6

Q.32

Find the real value of θ such that 3+2isinθ12isinθ is purely imaginary.

Ans

First, we shall put 3+2isinθ12isinθ in A+iB formWe have,3+2isinθ12isinθ=(3+2isinθ)(12isinθ)×(1+2isinθ)(1+2isinθ)               =3+2isinθ+6isinθ+4i2sin2θ14i2sin2θ               =3+8isinθ4sin2θ1+4sin2θ               =34sin2θ+i8sinθ1+4sin2θ               =34sin2θ1+4sin2θ+i8sinθ1+4sin2θFor complex number to be purely imaginary Real part=034sin2θ1+4sin2θ=0  34sin2θ=0       4sin2θ=3         sin2θ=34          sinθ=±32caseI             sinθ=32          sinθ=sinπ3              θ=π3caseII             sinθ=32          sinθ=sinπ3          sinθ=sinπ+π3          sinθ=sin4π3               θ=4π3

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FAQs (Frequently Asked Questions)

1. What are real numbers?

Real numbers are rational numbers that appear over the number line. The number line contains both rational and irrational numbers, numbering into the millions and billions. They could be either unfavourable or favourable. Rational numbers are indicated by the letter “R.” Rational numbers also include whole numbers, natural numbers, and integers. For instance, numbers beginning with 1 are called natural ones, while numbers beginning with 0 are called whole numbers.

2. Explain imaginary numbers.

The  complex numbers that are written in the format of real numbers are called imaginary numbers. They can be multiplied with an unreal or imaginary unit denoted as i. They are added to real numbers to form complex numbers in the form of x+yi where x and y are real numbers. However, here yi is the imaginary part of the complex number and x is the real part of the complex number.