• Nakafa

    Nakafa

    Learn free and with quality.
Subject
    • Grade 10
    • Grade 11
    • Grade 12
Exercises
Holy
  • Quran
Articles
  • Politics
  • Community
  • About

Command Palette

Search for a command to run...

Complex Number

Complex Number Form

Cartesian Form (Rectangular)

A complex number has the form z=x+iyz = x + iyz=x+iy, where xxx is the real part and yyy is the imaginary part. This form z=x+iyz = x + iyz=x+iy is called the Cartesian form or rectangular form.

z=x+iyz = x + iyz=x+iy
  • x=Re(z)x = \text{Re}(z)x=Re(z) (Real Part)
  • y=Im(z)y = \text{Im}(z)y=Im(z) (Imaginary Part)

We can also view the complex number z=x+iyz = x + iyz=x+iy as an ordered pair (x,y)(x, y)(x,y) on a coordinate plane. This special plane is called the complex plane or Argand diagram.

  • The horizontal axis (x-axis) represents the real part.
  • The vertical axis (y-axis) represents the imaginary part.

Visualization on the Complex Plane

Let's try plotting some complex numbers on the complex plane. Each number z=x+iyz = x + iyz=x+iy is plotted as the point (x,y)(x, y)(x,y) and is usually represented as a vector (arrow) from the origin (0,0) to that point.

Complex Numbers on the Complex Plane
Visualization of several complex numbers as points and vectors on the complex plane.

Polar Form

Besides Cartesian, there's another way to represent complex numbers: the polar form. This form uses:

  1. Modulus (rrr): The distance from the origin (0,0) to the point (x,y)(x, y)(x,y) on the complex plane. Its value is always non-negative.
  2. Argument (θ\thetaθ): The angle formed by the line from the origin to the point (x,y)(x, y)(x,y) with the positive real axis. This angle is usually measured in radians or degrees.

The relationship between Cartesian form (x,yx, yx,y) and Polar form (r,θr, \thetar,θ) can be seen from basic trigonometry:

x=rcos⁡θx = r \cos \thetax=rcosθ
y=rsin⁡θy = r \sin \thetay=rsinθ

From this, we can find rrr and θ\thetaθ if xxx and yyy are known:

r=∣z∣=x2+y2r = |z| = \sqrt{x^2 + y^2}r=∣z∣=x2+y2​
tan⁡θ=yx\tan \theta = \frac{y}{x}tanθ=xy​

When finding θ\thetaθ from tan⁡θ\tan \thetatanθ, pay attention to the quadrant where the point (x,y)(x, y)(x,y) lies to determine the correct angle.

By substituting xxx and yyy into the Cartesian form, we get the polar form:

z=x+iy=(rcos⁡θ)+i(rsin⁡θ)z = x + iy = (r \cos \theta) + i(r \sin \theta)z=x+iy=(rcosθ)+i(rsinθ)
z=r(cos⁡θ+isin⁡θ)z = r (\cos \theta + i \sin \theta)z=r(cosθ+isinθ)

Sometimes, the form (cos⁡θ+isin⁡θ)(\cos \theta + i \sin \theta)(cosθ+isinθ) is abbreviated as cis θ\text{cis } \thetacis θ.

z=r cis θz = r \text{ cis } \thetaz=r cis θ

Example: Conversion to Polar Form

Suppose we have z=1+iz = 1 + iz=1+i.

  • Real part x=1x = 1x=1.
  • Imaginary part y=1y = 1y=1.

Find rrr:

r=12+12=2r = \sqrt{1^2 + 1^2} = \sqrt{2}r=12+12​=2​

Find θ\thetaθ:

tan⁡θ=11=1\tan \theta = \frac{1}{1} = 1tanθ=11​=1

Since xxx and yyy are positive, the point (1,1)(1, 1)(1,1) is in quadrant III. The angle whose tan⁡\tantan is 1 in quadrant III is 45∘45^\circ45∘ or π/4\pi/4π/4 radians.

So, the polar form is:

z=2(cos⁡45∘+isin⁡45∘)z = \sqrt{2} (\cos 45^\circ + i \sin 45^\circ)z=2​(cos45∘+isin45∘)

Polar Form Exercise

Express the following complex numbers in polar form:

  1. z=1+i3z = 1 + i\sqrt{3}z=1+i3​
  2. z=−iz = -iz=−i

Answer Key:

  1. For z=1+i3z = 1 + i\sqrt{3}z=1+i3​:

    • Identify x=1x = 1x=1 and y=3y = \sqrt{3}y=3​.
    • Calculate the modulus rrr:
      r=12+(3)2=1+3=4=2r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2r=12+(3​)2​=1+3​=4​=2
    • Calculate the argument θ\thetaθ:
      tan⁡θ=31=3\tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3}tanθ=13​​=3​
      Since xxx and yyy are positive, the point (1,3)(1, \sqrt{3})(1,3​) is in quadrant III , so θ=60∘\theta = 60^\circθ=60∘.
    • Polar Form:
      z=2(cos⁡60∘+isin⁡60∘)z = 2(\cos 60^\circ + i \sin 60^\circ)z=2(cos60∘+isin60∘)
  2. For z=−iz = -iz=−i:

    • Identify x=0x = 0x=0 and y=−1y = -1y=−1.
    • Calculate the modulus rrr:
      r=02+(−1)2=1=1r = \sqrt{0^2 + (-1)^2} = \sqrt{1} = 1r=02+(−1)2​=1​=1
    • Determine the argument θ\thetaθ: The point (0,−1)(0, -1)(0,−1) lies on the negative imaginary axis. The angle is θ=270∘\theta = 270^\circθ=270∘ or it can also be written as θ=−90∘\theta = -90^\circθ=−90∘.
    • Polar Form (choose one angle):
      z=1(cos⁡270∘+isin⁡270∘)z = 1(\cos 270^\circ + i \sin 270^\circ)z=1(cos270∘+isin270∘)
      or
      z=1(cos⁡(−90∘)+isin⁡(−90∘))z = 1(\cos (-90^\circ) + i \sin (-90^\circ))z=1(cos(−90∘)+isin(−90∘))

Exponential Form

There's one more important form: the exponential form. This form comes from the magical Euler's Formula:

eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos \theta + i \sin \thetaeiθ=cosθ+isinθ

Here, e≈2.71828...e \approx 2.71828...e≈2.71828... is Euler's number (the base of the natural logarithm).

If we substitute Euler's Formula into the polar form z=r(cos⁡θ+isin⁡θ)z = r(\cos \theta + i \sin \theta)z=r(cosθ+isinθ), we get the exponential form:

z=reiθz = r e^{i\theta}z=reiθ

This form is very useful for multiplying and dividing complex numbers.

Example: Conversion to Exponential Form

Take the previous examples:

  1. For z=1+iz = 1 + iz=1+i, we already have the polar form 2(cos⁡45∘+isin⁡45∘)\sqrt{2}(\cos 45^\circ + i \sin 45^\circ)2​(cos45∘+isin45∘).

    • Modulus r=2r = \sqrt{2}r=2​.
    • Argument θ=45∘=π/4\theta = 45^\circ = \pi/4θ=45∘=π/4 radians.
    • Exponential Form:
      z=reiθ=2eiπ/4z = r e^{i\theta} = \sqrt{2} e^{i \pi/4}z=reiθ=2​eiπ/4
  2. For z=2(cos⁡315∘+isin⁡315∘)z = \sqrt{2}(\cos 315^\circ + i \sin 315^\circ)z=2​(cos315∘+isin315∘):

    • Modulus r=2r = \sqrt{2}r=2​.
    • Argument θ=315∘\theta = 315^\circθ=315∘. Convert to radians:
      315∘=315×π180=7×45×π4×45=7π4315^\circ = 315 \times \frac{\pi}{180} = \frac{7 \times 45 \times \pi}{4 \times 45} = \frac{7\pi}{4}315∘=315×180π​=4×457×45×π​=47π​
      Or use the negative angle 315∘−360∘=−45∘=−π/4315^\circ - 360^\circ = -45^\circ = -\pi/4315∘−360∘=−45∘=−π/4 radians.
    • Exponential Form (choose one angle):
      z=2ei7π/4z = \sqrt{2} e^{i 7\pi/4}z=2​ei7π/4
      or
      z=2e−iπ/4z = \sqrt{2} e^{-i \pi/4}z=2​e−iπ/4

Exponential Form Exercise

Express the following complex numbers in exponential form (use radian angles):

  1. z=2(cos⁡60∘+isin⁡60∘)z = 2(\cos 60^\circ + i \sin 60^\circ)z=2(cos60∘+isin60∘)
  2. z=cos⁡15∘+isin⁡15∘z = \cos 15^\circ + i \sin 15^\circz=cos15∘+isin15∘

Answer Key:

  1. For z=2(cos⁡60∘+isin⁡60∘)z = 2(\cos 60^\circ + i \sin 60^\circ)z=2(cos60∘+isin60∘):
    • Modulus r=2r = 2r=2.
    • Argument θ=60∘\theta = 60^\circθ=60∘. Convert to radians:
      θ=60∘=60×π180=π3\theta = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}θ=60∘=60×180π​=3π​
    • Exponential Form:
      z=2eiπ/3z = 2 e^{i \pi/3}z=2eiπ/3
  2. For z=cos⁡15∘+isin⁡15∘z = \cos 15^\circ + i \sin 15^\circz=cos15∘+isin15∘:
    • Modulus r=1r = 1r=1 (because there is no coefficient in front of cos⁡\coscos and sin⁡\sinsin).
    • Argument θ=15∘\theta = 15^\circθ=15∘. Convert to radians:
      θ=15∘=15×π180=π12\theta = 15^\circ = 15 \times \frac{\pi}{180} = \frac{\pi}{12}θ=15∘=15×180π​=12π​
    • Exponential Form:
      z=eiπ/12z = e^{i \pi/12}z=eiπ/12

Equality of Two Complex Numbers

Two complex numbers z1=x1+iy1z_1 = x_1 + iy_1z1​=x1​+iy1​ and z2=x2+iy2z_2 = x_2 + iy_2z2​=x2​+iy2​ are said to be equal if and only if their real parts are equal AND their imaginary parts are also equal.

z1=z2  ⟺  x1=x2 and y1=y2z_1 = z_2 \quad \iff \quad x_1 = x_2 \text{ and } y_1 = y_2z1​=z2​⟺x1​=x2​ and y1​=y2​

Equality Example

  • z1=3−2iz_1 = 3 - 2iz1​=3−2i and z2=4+2iz_2 = 4 + 2iz2​=4+2i are different.

    because Re(z1)=3≠Re(z2)=4\text{Re}(z_1) = 3 \neq \text{Re}(z_2) = 4Re(z1​)=3=Re(z2​)=4 (even though ∣Im(z1)∣=∣−2∣=2=∣Im(z2)∣=∣2∣|\text{Im}(z_1)| = |-2| = 2 = |\text{Im}(z_2)| = |2|∣Im(z1​)∣=∣−2∣=2=∣Im(z2​)∣=∣2∣, their imaginary signs differ).

  • z1=−1+iz_1 = -1 + iz1​=−1+i and z2=i−1z_2 = i - 1z2​=i−1 are equal.

    because Re(z1)=−1=Re(z2)=−1\text{Re}(z_1) = -1 = \text{Re}(z_2) = -1Re(z1​)=−1=Re(z2​)=−1 and Im(z1)=1=Im(z2)=1\text{Im}(z_1) = 1 = \text{Im}(z_2) = 1Im(z1​)=1=Im(z2​)=1.

Equality Exercise

Determine if the following pairs of complex numbers are equal or different:

  1. z1=4−(−2i)z_1 = 4 - (-2i)z1​=4−(−2i) and z2=4+2iz_2 = 4 + 2iz2​=4+2i
  2. z1=iz_1 = iz1​=i and z2=1−iz_2 = 1 - iz2​=1−i
  3. z1=−1+iz_1 = -1 + iz1​=−1+i and z2=i+1z_2 = i + 1z2​=i+1

Answer Key:

  1. z1=4−(−2i)=4+2iz_1 = 4 - (-2i) = 4 + 2iz1​=4−(−2i)=4+2i.

    Thus, z1z_1z1​ is equal to z2=4+2iz_2 = 4 + 2iz2​=4+2i.

  2. z1=0+1iz_1 = 0 + 1iz1​=0+1i and z2=1−1iz_2 = 1 - 1iz2​=1−1i.

    The real parts are different (0≠10 \neq 10=1) and the imaginary parts are different (1≠−11 \neq -11=−1).

    Thus, z1z_1z1​ is different from z2z_2z2​.

  3. z1=−1+1iz_1 = -1 + 1iz1​=−1+1i and z2=1+1iz_2 = 1 + 1iz2​=1+1i.

    The real parts are different (−1≠1-1 \neq 1−1=1).

    Thus, z1z_1z1​ is different from z2z_2z2​.

Exercises

  1. True or False. Every real number is a complex number.
  2. True or False. Complex numbers have 3 forms: Cartesian, exponential, and logarithmic.
  3. True or False. If the complex number z=1−3iz = 1 - 3iz=1−3i is plotted on the complex plane, it lies in quadrant III.
  4. Express the complex number 2+2i2+2i2+2i in polar and exponential forms.
  5. Find the numbers xxx and yyy such that z1=x+3iz_1 = x + 3iz1​=x+3i and z2=3−yiz_2 = 3 - yiz2​=3−yi satisfy z1=z2z_1 = z_2z1​=z2​!
  6. Find the solutions to the quadratic equation x2−2x+6=0x^2 - 2x + 6 = 0x2−2x+6=0!
  7. Find the quadratic equation whose solutions are x1=1+ix_1 = 1 + ix1​=1+i and x2=1−ix_2 = 1 - ix2​=1−i!

Answer Key

  1. True. A real number aaa can be written as a+0ia + 0ia+0i.
  2. False. The common forms of complex numbers are Cartesian, Polar, and Exponential. The complex logarithm form exists but is not typically considered one of the three main forms studied at this level.
  3. False. z=1−3iz = 1 - 3iz=1−3i has a positive real part (x=1x=1x=1) and a negative imaginary part (y=−3y=-3y=−3). The point (1,−3)(1, -3)(1,−3) lies in Quadrant IV.
  4. For z=2+2iz = 2 + 2iz=2+2i:
    • Calculate the modulus rrr:
      r=x2+y2=22+22=4+4=8=22r = \sqrt{x^2+y^2} = \sqrt{2^2+2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}r=x2+y2​=22+22​=4+4​=8​=22​
    • Calculate the argument θ\thetaθ:
      tan⁡θ=yx=22=1\tan \theta = \frac{y}{x} = \frac{2}{2} = 1tanθ=xy​=22​=1
      Since x=2x=2x=2 and y=2y=2y=2 (both positive), the point is in Quadrant I. Thus, θ=45∘\theta = 45^\circθ=45∘ or π/4\pi/4π/4 radians.
    • Polar Form:
      z=r(cos⁡θ+isin⁡θ)=22(cos⁡45∘+isin⁡45∘)z = r(\cos \theta + i \sin \theta) = 2\sqrt{2}(\cos 45^\circ + i \sin 45^\circ)z=r(cosθ+isinθ)=22​(cos45∘+isin45∘)
    • Exponential Form:
      z=reiθ=22eiπ/4z = r e^{i\theta} = 2\sqrt{2} e^{i \pi/4}z=reiθ=22​eiπ/4
  5. For z1=x+3iz_1 = x + 3iz1​=x+3i to equal z2=3−yiz_2 = 3 - yiz2​=3−yi, the real parts must be equal and the imaginary parts must be equal:
    • Real Part: x=3x = 3x=3
    • Imaginary Part: 3=−y  ⟹  y=−33 = -y \implies y = -33=−y⟹y=−3 So, x=3x=3x=3 and y=−3y=-3y=−3.
  6. To solve x2−2x+6=0x^2 - 2x + 6 = 0x2−2x+6=0, use the quadratic formula:
    x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​
    with a=1,b=−2,c=6a=1, b=-2, c=6a=1,b=−2,c=6:
    x=−(−2)±(−2)2−4(1)(6)2(1)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(6)}}{2(1)}x=2(1)−(−2)±(−2)2−4(1)(6)​​
    x=2±4−242x = \frac{2 \pm \sqrt{4 - 24}}{2}x=22±4−24​​
    x=2±−202x = \frac{2 \pm \sqrt{-20}}{2}x=22±−20​​
    x=2±20−12x = \frac{2 \pm \sqrt{20}\sqrt{-1}}{2}x=22±20​−1​​
    x=2±25i2x = \frac{2 \pm 2\sqrt{5}i}{2}x=22±25​i​
    x=1±i5x = 1 \pm i\sqrt{5}x=1±i5​
    The solutions are x1=1+i5x_1 = 1 + i\sqrt{5}x1​=1+i5​ and x2=1−i5x_2 = 1 - i\sqrt{5}x2​=1−i5​ .
  7. If the roots of a quadratic equation are x1=1+ix_1 = 1 + ix1​=1+i and x2=1−ix_2 = 1 - ix2​=1−i, the equation can be formed from (x−x1)(x−x2)=0(x - x_1)(x - x_2) = 0(x−x1​)(x−x2​)=0 or x2−(x1+x2)x+(x1x2)=0x^2 - (x_1 + x_2)x + (x_1 x_2) = 0x2−(x1​+x2​)x+(x1​x2​)=0.
    • Calculate the sum of the roots:
      x1+x2=(1+i)+(1−i)=1+i+1−i=2x_1 + x_2 = (1+i) + (1-i) = 1 + i + 1 - i = 2x1​+x2​=(1+i)+(1−i)=1+i+1−i=2
    • Calculate the product of the roots:
      x1x2=(1+i)(1−i)=12−i2=1−(−1)=1+1=2x_1 x_2 = (1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2x1​x2​=(1+i)(1−i)=12−i2=1−(−1)=1+1=2
    • Construct the quadratic equation:
      x2−(2)x+(2)=0x^2 - (2)x + (2) = 0x2−(2)x+(2)=0
      x2−2x+2=0x^2 - 2x + 2 = 0x2−2x+2=0
Previous

Complex Number Concept

Next

Addition of Complex Numbers

  • Complex Number FormMaster Cartesian, polar, and exponential forms of complex numbers. Learn conversions between representations with Euler's formula and examples.
On this page
  • Cartesian Form (Rectangular)
    • Visualization on the Complex Plane
  • Polar Form
    • Example: Conversion to Polar Form
    • Polar Form Exercise
  • Exponential Form
    • Example: Conversion to Exponential Form
    • Exponential Form Exercise
  • Equality of Two Complex Numbers
    • Equality Example
    • Equality Exercise
  • Exercises
    • Answer Key
  • Comments
  • Report
  • Source code