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Complex Number

Complex Number Form

Cartesian Form (Rectangular)

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

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

We can also view the complex number z=x+iyz = x + iy as an ordered pair (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 + iy is plotted as the point (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 (rr): The distance from the origin (0,0) to the point (x,y)(x, y) on the complex plane. Its value is always non-negative.
  2. Argument (θ\theta): The angle formed by the line from the origin to the point (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, y) and Polar form (r,θr, \theta) can be seen from basic trigonometry:

x=rcosθx = r \cos \theta
y=rsinθy = r \sin \theta

From this, we can find rr and θ\theta if xx and yy are known:

r=z=x2+y2r = |z| = \sqrt{x^2 + y^2}
tanθ=yx\tan \theta = \frac{y}{x}

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

By substituting xx and yy into the Cartesian form, we get the polar form:

z=x+iy=(rcosθ)+i(rsinθ)z = x + iy = (r \cos \theta) + i(r \sin \theta)
z=r(cosθ+isinθ)z = r (\cos \theta + i \sin \theta)

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

z=r cis θz = r \text{ cis } \theta

Example: Conversion to Polar Form

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

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

Find rr:

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

Find θ\theta:

tanθ=11=1\tan \theta = \frac{1}{1} = 1

Since xx and yy are positive, the point (1,1)(1, 1) is in quadrant II. The angle whose tan\tan is 1 in quadrant II is 4545^\circ or π/4\pi/4 radians.

So, the polar form is:

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

Polar Form Exercise

Express the following complex numbers in polar form:

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

Answer Key:

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

    • Identify x=1x = 1 and y=3y = \sqrt{3}.
    • Calculate the modulus rr:
      r=12+(3)2=1+3=4=2r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
    • Calculate the argument θ\theta:
      tanθ=31=3\tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3}
      Since xx and yy are positive, the point (1,3)(1, \sqrt{3}) is in quadrant II , so θ=60\theta = 60^\circ.
    • Polar Form:
      z=2(cos60+isin60)z = 2(\cos 60^\circ + i \sin 60^\circ)
  2. For z=iz = -i:

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

Exponential Form

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

eiθ=cosθ+isinθe^{i\theta} = \cos \theta + i \sin \theta

Here, e2.71828...e \approx 2.71828... is Euler's number (the base of the natural logarithm).

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

z=reiθz = r e^{i\theta}

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 + i, we already have the polar form 2(cos45+isin45)\sqrt{2}(\cos 45^\circ + i \sin 45^\circ).

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

    • Modulus r=2r = \sqrt{2}.
    • Argument θ=315\theta = 315^\circ. 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}
      Or use the negative angle 315360=45=π/4315^\circ - 360^\circ = -45^\circ = -\pi/4 radians.
    • Exponential Form (choose one angle):
      z=2ei7π/4z = \sqrt{2} e^{i 7\pi/4}
      or
      z=2eiπ/4z = \sqrt{2} e^{-i \pi/4}

Exponential Form Exercise

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

  1. z=2(cos60+isin60)z = 2(\cos 60^\circ + i \sin 60^\circ)
  2. z=cos15+isin15z = \cos 15^\circ + i \sin 15^\circ

Answer Key:

  1. For z=2(cos60+isin60)z = 2(\cos 60^\circ + i \sin 60^\circ):
    • Modulus r=2r = 2.
    • Argument θ=60\theta = 60^\circ. Convert to radians:
      θ=60=60×π180=π3\theta = 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3}
    • Exponential Form:
      z=2eiπ/3z = 2 e^{i \pi/3}
  2. For z=cos15+isin15z = \cos 15^\circ + i \sin 15^\circ:
    • Modulus r=1r = 1 (because there is no coefficient in front of cos\cos and sin\sin).
    • Argument θ=15\theta = 15^\circ. Convert to radians:
      θ=15=15×π180=π12\theta = 15^\circ = 15 \times \frac{\pi}{180} = \frac{\pi}{12}
    • Exponential Form:
      z=eiπ/12z = e^{i \pi/12}

Equality of Two Complex Numbers

Two complex numbers z1=x1+iy1z_1 = x_1 + iy_1 and z2=x2+iy2z_2 = x_2 + iy_2 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_2

Equality Example

  • z1=32iz_1 = 3 - 2i and z2=4+2iz_2 = 4 + 2i are different.

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

  • z1=1+iz_1 = -1 + i and z2=i1z_2 = i - 1 are equal.

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

Equality Exercise

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

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

Answer Key:

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

    Thus, z1z_1 is equal to z2=4+2iz_2 = 4 + 2i.

  2. z1=0+1iz_1 = 0 + 1i and z2=11iz_2 = 1 - 1i.

    The real parts are different (010 \neq 1) and the imaginary parts are different (111 \neq -1).

    Thus, z1z_1 is different from z2z_2.

  3. z1=1+1iz_1 = -1 + 1i and z2=1+1iz_2 = 1 + 1i.

    The real parts are different (11-1 \neq 1).

    Thus, z1z_1 is different from z2z_2.

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=13iz = 1 - 3i is plotted on the complex plane, it lies in quadrant III.
  4. Express the complex number 2+2i2+2i in polar and exponential forms.
  5. Find the numbers xx and yy such that z1=x+3iz_1 = x + 3i and z2=3yiz_2 = 3 - yi satisfy z1=z2z_1 = z_2!
  6. Find the solutions to the quadratic equation x22x+6=0x^2 - 2x + 6 = 0!
  7. Find the quadratic equation whose solutions are x1=1+ix_1 = 1 + i and x2=1ix_2 = 1 - i!

Answer Key

  1. True. A real number aa can be written as a+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=13iz = 1 - 3i has a positive real part (x=1x=1) and a negative imaginary part (y=3y=-3). The point (1,3)(1, -3) lies in Quadrant IV.
  4. For z=2+2iz = 2 + 2i:
    • Calculate the modulus rr:
      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}
    • Calculate the argument θ\theta:
      tanθ=yx=22=1\tan \theta = \frac{y}{x} = \frac{2}{2} = 1
      Since x=2x=2 and y=2y=2 (both positive), the point is in Quadrant I. Thus, θ=45\theta = 45^\circ or π/4\pi/4 radians.
    • Polar Form:
      z=r(cosθ+isinθ)=22(cos45+isin45)z = r(\cos \theta + i \sin \theta) = 2\sqrt{2}(\cos 45^\circ + i \sin 45^\circ)
    • Exponential Form:
      z=reiθ=22eiπ/4z = r e^{i\theta} = 2\sqrt{2} e^{i \pi/4}
  5. For z1=x+3iz_1 = x + 3i to equal z2=3yiz_2 = 3 - yi, the real parts must be equal and the imaginary parts must be equal:
    • Real Part: x=3x = 3
    • Imaginary Part: 3=y    y=33 = -y \implies y = -3 So, x=3x=3 and y=3y=-3.
  6. To solve x22x+6=0x^2 - 2x + 6 = 0, use the quadratic formula:
    x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
    with a=1,b=2,c=6a=1, b=-2, c=6:
    x=(2)±(2)24(1)(6)2(1)x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(6)}}{2(1)}
    x=2±4242x = \frac{2 \pm \sqrt{4 - 24}}{2}
    x=2±202x = \frac{2 \pm \sqrt{-20}}{2}
    x=2±2012x = \frac{2 \pm \sqrt{20}\sqrt{-1}}{2}
    x=2±25i2x = \frac{2 \pm 2\sqrt{5}i}{2}
    x=1±i5x = 1 \pm i\sqrt{5}
    The solutions are x1=1+i5x_1 = 1 + i\sqrt{5} and x2=1i5x_2 = 1 - i\sqrt{5} .
  7. If the roots of a quadratic equation are x1=1+ix_1 = 1 + i and x2=1ix_2 = 1 - i, the equation can be formed from (xx1)(xx2)=0(x - x_1)(x - x_2) = 0 or x2(x1+x2)x+(x1x2)=0x^2 - (x_1 + x_2)x + (x_1 x_2) = 0.
    • Calculate the sum of the roots:
      x1+x2=(1+i)+(1i)=1+i+1i=2x_1 + x_2 = (1+i) + (1-i) = 1 + i + 1 - i = 2
    • Calculate the product of the roots:
      x1x2=(1+i)(1i)=12i2=1(1)=1+1=2x_1 x_2 = (1+i)(1-i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2
    • Construct the quadratic equation:
      x2(2)x+(2)=0x^2 - (2)x + (2) = 0
      x22x+2=0x^2 - 2x + 2 = 0