Mada za sehemu hiiComplex NumbersMada 4
- Complex Numbers And Their Operations
- Polar Form Of Complex Numbers
- De Moivre’s theorem
- Euler’s formula
Consider a general quadratic equation ax2+bx+c=0, where a=0. The general solution is given by:
x=2a−b±b2−4ac
The quadratic equation has no solution in the set of real numbers if the discriminant is negative, i.e., if b2−4ac<0.
For example, consider x2+2x+3=0. Using the quadratic formula:
x=2(1)−2±22−4(1)(3)=2−2±4−12=2−2±−8=2−2±8×−1=2−2±22−1=−1±2−1
Since the square of any real number is positive, −1 is not a real number. −1 is defined as the imaginary unit, denoted by i, where i=−1 or i2=−1.
Solutions of this type are found in the set of complex numbers. A complex number has a real and an imaginary part. It is expressed in the form x+iy, where x and y are real numbers and i is the imaginary unit.
In x+iy, x is the real part and y is the imaginary part. The solution to x2+2x+3=0 is x=−1±i2, so x=−1+i2 or x=−1−i2.
The set of complex numbers is denoted by C. A member of C can be denoted by a complex variable z=x+iy. The real part of z is denoted as Re(z) = x, and the imaginary part is denoted as Im(z) = y. If x=0, z is purely imaginary. If y=0, z is purely real.
A complex number z=x+iy can be represented geometrically in the complex plane, called the Argand diagram. In the Argand diagram, z=x+iy is represented by the point P(x, y), where Re(z) is on the x-axis (real axis) and Im(z) is on the y-axis (imaginary axis).
The Argand diagram
If the discriminant in the general solution of the quadratic equation ax2+bx+c=0 is negative, then the solution lies in the set of complex numbers.
Example 1
Find the solution of the equation x2+2x+2=0.
Solution:
Using the quadratic formula, x=2a−b±b2−4ac, with a=1, b=2, and c=2:
x=2(1)−2±22−4(1)(2)=2−2±4−8=2−2±−4
Since −4=4×−1=4×−1=2i, where i=−1:
x=2−2±2i=−1±i
Therefore, the solutions are x=−1+i and x=−1−i.
Example 2
Solve for x in the equation x2−4x+7=0.
Solution:
Using the quadratic formula, x=2a−b±b2−4ac, with a=1, b=−4, and c=7:
x=2(1)−(−4)±(−4)2−4(1)(7)=24±16−28=24±−12
Since −12=12×−1=4×3×−1=23i:
x=24±23i=2±i3
Therefore, the solutions are x=2+i3 and x=2−i3.
The powers of the imaginary unit i are essential for multiplying and simplifying complex numbers.
Recall that i=−1, so i2=−1.
If i is raised to a non-negative integer n, the values cycle through a pattern:
(a) i3=i2×i=(−1)i=−i
(b) i4=i2×i2=(−1)(−1)=1
(c) i5=i4×i=(1)i=i
(d) i6=i4×i2=(1)(−1)=−1
(e) i7=i6×i=(−1)i=−i
The pattern repeats every four powers: i,−1,−i,1.
Example 3
Simplify each of the following complex numbers:
(a) i1776 (b) i13+i19 (c) i5+i3+i+1 (d) −48−−27
Solution:
(a) i1776. Since 1776 is divisible by 4 (1776=4×444), i1776=(i4)444=1444=1.
(b) i13+i19. i13=i12×i=(i4)3×i=13×i=i. i19=i16×i3=(i4)4×i3=14×−i=−i. Therefore, i13+i19=i+(−i)=0.
(c) i5+i3+i+1. i5=i, i3=−i. Therefore, i5+i3+i+1=i+(−i)+i+1=i+1.
(d) −48−−27. −48=16×3×−1=4i3. −27=9×3×−1=3i3. Therefore, −48−−27=4i3−3i3=i3.
Example 4
Find the complex number z in the equation 4+5i=z−(1−i).
Solution:
z=(4+5i)+(1−i)=4+1+5i−i=5+4i
Example 5
Plot the complex numbers z=1+2i and w=1−2i on the same Argand diagram. (Argand diagram not included, but would show points (1,2) and (1,-2))
Solution:
The Argand diagram for the given complex numbers is as shown in the following figure.
Example 6
Find the imaginary and real parts in each of the following complex numbers:
(a) (3+i)(3−i) (b) (2+i)2 (c) (−2−3i)3 (d) 2+3i2−i
Solution:
(a) (3+i)(3−i)=9−i2=9−(−1)=10. Re(z) = 10, Im(z) = 0.
(b) (2+i)2=4+4i+i2=4+4i−1=3+4i. Re(w) = 3, Im(w) = 4.
(c) (−2−3i)3=(−1)3(2+3i)3=−(8+36i+54i2+27i3)=−(8+36i−54−27i)=−(−46+9i)=46−9i. Re(z) = 46, Im(z) = -9.
(d) 2+3i2−i=(2+3i)(2−3i)(2−i)(2−3i)=4−9i24−6i−2i+3i2=4+94−8i−3=131−8i=131−138i. Re(z) = 131, Im(z) = −138.
If z=x+iy is a complex number, the modulus of z, denoted by ∣z∣ or Mod(z), is the distance of the point representing z on the Argand plane from the origin. It is calculated as:
∣z∣=x2+y2
Geometrical representation of a complex number
Basic properties of the modulus:
- ∣z∣=0 if and only if z=0 (i.e., Re(z) = 0 and Im(z) = 0)
- -∣z∣ ≤ Re(z) ≤ ∣z∣ and -∣z∣ ≤ Im(z) ≤ ∣z∣
- ∣z1z2∣=∣z1∣∣z2∣
- ∣z2z1∣=∣z2∣∣z1∣, z2=0
- ∣z1+z2∣≤∣z1∣+∣z2∣ (Triangle Inequality)
- ∣z1−z2∣≥∣∣z1∣−∣z2∣∣
- ∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)
Example 7
Find the modulus of each complex number:
(a) z=−3+i (b) w=2+7i (c) u=5+2i
Solution:
(a) ∣z∣=(−3)2+12=9+1=10
(b) ∣w∣=22+72=4+49=53
(c) ∣u∣=52+22=25+4=29
Example 8
Find the modulus of z if:
(a) z=−26−3i (b) z=(a+b)+i(a−b)
Solution:
(a) ∣z∣=(−26)2+(−3)2=24+9=33
(b) ∣z∣=(a+b)2+(a−b)2=a2+2ab+b2+a2−2ab+b2=2a2+2b2=2(a2+b2)
Example 9
If z1=2+3i and z2=−3+4i, verify that ∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2).
Solution:
Left-hand side:
z1+z2=(2+3i)+(−3+4i)=−1+7i. ∣z1+z2∣2=(−1)2+72=1+49=50
z1−z2=(2+3i)−(−3+4i)=5−i. ∣z1−z2∣2=52+(−1)2=25+1=26
∣z1+z2∣2+∣z1−z2∣2=50+26=76
Right-hand side:
∣z1∣2=22+32=4+9=13
∣z2∣2=(−3)2+42=9+16=25
2(∣z1∣2+∣z2∣2)=2(13+25)=2(38)=76
Since both sides equal 76, the property is verified.
For a complex number z=x+iy, the complex conjugate, denoted by zˉ, is x−iy. The conjugate is obtained by changing the sign of the imaginary part.
Basic properties of complex conjugates:
- zˉ=z if and only if z is a real number (i.e., z∈R).
- If z is a complex number, then zzˉ is a non-negative real number.
- The conjugate of a sum is the sum of the conjugates: z1+z2=z1ˉ+z2ˉ.
- The conjugate of a product is the product of the conjugates: z1z2=z1ˉz2ˉ.
- The conjugate of a quotient is the quotient of the conjugates: (z2z1)=z2ˉz1ˉ, for z2=0.
- Re(z) = 2z+zˉ and Im(z) = 2iz−zˉ.
- zzˉ=∣z∣2.
- The multiplicative inverse (reciprocal) of a complex number z=a+ib (b=0) is z1=∣z∣2zˉ=a2+b2a−ib.
Example 10
Find the complex conjugate of each complex number:
(a) 3+6i (b) 21−5i (c) a+ib
Solution:
(a) If z=3+6i, then zˉ=3−6i.
(b) If w=21−5i, then wˉ=21+5i.
(c) If z=a+ib, then zˉ=a−ib.
Example 11
Show that z1+z2=z1ˉ+z2ˉ.
Solution:
Let z1=x1+iy1 and z2=x2+iy2. Then:
z1+z2=(x1+iy1)+(x2+iy2)=(x1+x2)+i(y1+y2)
z1+z2=(x1+x2)−i(y1+y2)
Also:
z1ˉ=x1−iy1
z2ˉ=x2−iy2
z1ˉ+z2ˉ=(x1−iy1)+(x2−iy2)=(x1+x2)−i(y1+y2)
Therefore, z1+z2=z1ˉ+z2ˉ.
If P(x, y) is a point in an Argand diagram representing a complex number z=x+iy, the modulus r is the same as ∣z∣. The angle θ is called the argument of z and is denoted by arg(z). The principal argument lies in the range (−π,π] or (−180∘,180∘]. The argument is undefined when z=0.
Modulus and argument in Argand diagram
From Figure tan(θ)=xy, so θ=arctan(xy). However, care must be taken to choose the correct quadrant for θ based on the signs of x and y. The argument of a complex number z=x+iy is given by:
arg(z) = arctan(xy), with quadrant adjustments as needed.
Example 12
Find the argument of each complex number:
(a) z=2+3i (b) w=1+i (c) p=5−2i
Solution:
(a) z=2+3i. x=2, y=3. θ=arctan(23)≈56.3∘. Since x and y are positive, z is in the first quadrant, so arg(z) ≈56.3∘.
(b) w=1+i. x=1, y=1. θ=arctan(11)=arctan(1)=45∘. Since x and y are positive, w is in the first quadrant, so arg(w) = 45∘.
(c) p=5−2i. x=5, y=−2. θ=arctan(5−2)≈−21.8∘. Since x is positive and y is negative, p is in the fourth quadrant, so arg(p) ≈−21.8∘.
Example 13
If z1=−5+6i and z2=−1−2i, find the argument of z1+z2.
Solution:
z1+z2=(−5+6i)+(−1−2i)=−6+4i
Let z=z1+z2=−6+4i. Here x=−6 and y=4. Because x is negative and y is positive, the complex number lies in the second quadrant.
θ=arctan(−64)≈−33.7∘. This is the reference angle.
Since the complex number lies in the second quadrant, we need to add 180° to the reference angle:
arg(z) = -33.7° + 180° = 146.3°
Or in radians:
θ=arctan(−64)≈−0.59 radians.
Since it's in the second quadrant, we add π:
arg(z) = −0.59+π≈2.55 radians.
The fundamental operations on complex numbers are addition, subtraction, multiplication, and division. These operations are performed similarly to those on polynomials.
Given complex numbers z=a+ib and w=c+id:
Addition
Addition of z and w is done by adding the real and imaginary parts separately:
z+w=(a+ib)+(c+id)=(a+c)+i(b+d)
Subtraction
Subtraction is done by subtracting the real and imaginary parts separately:
z−w=(a+ib)−(c+id)=(a−c)+i(b−d)
Multiplication
Multiplication is similar to multiplying polynomial factors:
zw=(a+ib)(c+id)=ac+iad+ibc+i2bd=ac+i(ad+bc)−bd=(ac−bd)+i(ad+bc)
Division
Division is done by multiplying the numerator and denominator by the complex conjugate of the denominator:
wz=c+ida+ib=(c+id)(c−id)(a+ib)(c−id)=c2−i2d2ac−iad+ibc−i2bd=c2+d2ac+bd+i(bc−ad)=c2+d2ac+bd+ic2+d2bc−ad
Example 15
Simplify each of the following expressions:
(a) (3+4i)(−3−i) (b) (−5−10i)(3+i) (c) 3−i1+3i
Solution:
(a) (3+4i)(−3−i)=−9−3i−12i−4i2=−9−15i+4=−5−15i
(b) (−5−10i)(3+i)=−15−5i−30i−10i2=−15−35i+10=−5−35i
(c) 3−i1+3i=(3−i)(3+i)(1+3i)(3+i)=9−i23+i+9i+3i2=9+13+10i−3=1010i=i
Example 16
Solve for z and w in the following system of equations:
2z+iw=8 (i)
iz−w=4+2i (ii)
Solution:
Multiply equation (i) by 3 and equation (ii) by i:
6z+3iw=24 (iii)
−z−iw=4i−2 (iv)
Add equations (iii) and (iv):
5z=22+4i
z=522+54i
Substitute the value of z into equation (i):
2(522+54i)+iw=8
544+58i+iw=8
iw=8−544−58i
iw=540−44−58i
iw=−54−58i
w=−54i−1−58=54i−58
w=−58+54i
Therefore, z=522+54i and w=−58+54i.
Example 17
Given z=3+i and w=3i, determine the simplified expressions for zw and wz, giving the answers in the form x+iy, where x,y∈R.
Solution:
zw=(3+i)(3i)=9i+3i2=9i−3=−3+9i
wz=3i3+i=(3i)(−3i)(3+i)(−3i)=−9i2−9i−3i2=9−9i+3=31−i
Equations can be solved using different methods, including factorization and the general quadratic formula. Similarly, solutions of equations involving complex numbers can be obtained by these methods.
If z is a complex number, the general polynomial equation involving complex numbers takes the form:
f(z)=anzn+an−1zn−1+...+a1z+a0=0, where a0,a1,...,an are real coefficients.
If z is a root of the equation f(z)=0, then the complex conjugate of z is also a root of the equation. This means that complex roots of polynomial equations with real coefficients occur in complex conjugate pairs.
Example 18
Find the solution of the polynomial equation z3−z2+25z−25=0.
Solution:
Let f(z)=z3−z2+25z−25. The constant term is -25. Factors of 25 are ±1, ±5, and ±25.
Testing these values:
f(1)=1−1+25−25=0. Thus, z=1 is a root.
Since z=1 is a root, z−1 is a factor. Dividing the polynomial by z−1 gives:
z−1z3−z2+25z−25=z2+25
The quadratic equation z2+25=0 can be solved as follows:
z2=−25
z=±−25=±5i
Therefore, the solutions are z=1, z=5i, and z=−5i.
Example 19
Find the roots of the polynomial equation 9z4−35z2+4=0.
Solution:
Let u=z2. The equation becomes:
9u2−35u+4=0
Using the quadratic formula:
u=2(9)35±(−35)2−4(9)(4)=1835±1225−144=1835±1081=1835±33
u=1868=934 or u=182=91
Since u=z2:
z2=934 or z2=91
z=±334 or z=±31
Therefore, the roots are z=334, z=−334, z=31, and z=−31.
Example 20
If 2+i is one of the roots of the equation z3−5z2+9z−5=0, find the other roots.
Solution:
Since 2+i is a root, its conjugate 2−i is also a root.
The quadratic factor corresponding to these roots is (z−(2+i))(z−(2−i))=(z−2−i)(z−2+i)=(z−2)2−i2=z2−4z+4+1=z2−4z+5.
Dividing the original cubic by this quadratic factor:
z2−4z+5z3−5z2+9z−5=z−1
Therefore, z−1=0 which means z=1 is the remaining root.
The other roots are 2−i and 1.
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