Math Problem Statement

Solve a series of complex number problems involving exponential polar form, modulus, principal argument, and Cartesian equations in the complex plane.

Solution

Let's work through the mathematical problems based on the uploaded image. Here's a step-by-step approach for each part.

Question 2 (a):

We need to express the complex number z=i(22i)(13i)2z = \frac{i(2 - 2i)}{(1 - \sqrt{3}i)^2} in exponential polar form.

Step 1: Simplify the expression

  • First, simplify the numerator: i(22i)=2i+2i(2 - 2i) = 2i + 2 So, the numerator becomes 2+2i2 + 2i.

  • Now, simplify the denominator: (13i)2=(13i)(13i)=122(1)(3i)+(3i)2(1 - \sqrt{3}i)^2 = (1 - \sqrt{3}i)(1 - \sqrt{3}i) = 1^2 - 2(1)(\sqrt{3}i) + (\sqrt{3}i)^2 =123i3=223i= 1 - 2\sqrt{3}i - 3 = -2 - 2\sqrt{3}i Therefore, z=2+2i223iz = \frac{2 + 2i}{-2 - 2\sqrt{3}i}.

Step 2: Multiply numerator and denominator by the conjugate

  • To remove the imaginary part from the denominator, multiply by the conjugate 2+23i-2 + 2\sqrt{3}i: z=(2+2i)(2+23i)(223i)(2+23i)z = \frac{(2 + 2i)(-2 + 2\sqrt{3}i)}{(-2 - 2\sqrt{3}i)(-2 + 2\sqrt{3}i)} Simplifying this results in a complex number that we can convert into polar form.

Step 3: Convert to polar form

  • We need to find the modulus rr and argument θ\theta of the simplified form to express the complex number in the exponential form z=reiθz = r e^{i\theta}.

Question 2 (b):

Given z=5eiπ/25z = 5e^{i\pi/25}, we are asked to find the modulus and the principal argument of z80z^{80}.

Step 1: Modulus of zz

  • The modulus of zz is simply the coefficient in front of the exponential, so z=5|z| = 5.
  • The modulus of z80z^{80} is z80=z80=580|z^{80}| = |z|^{80} = 5^{80}.

Step 2: Principal argument

  • The argument of zz is π25\frac{\pi}{25}, so the argument of z80z^{80} is: Arg(z80)=80×π25=80π25=16π5\text{Arg}(z^{80}) = 80 \times \frac{\pi}{25} = \frac{80\pi}{25} = \frac{16\pi}{5} Since principal arguments are typically restricted to the interval (π,π](-\pi, \pi], we need to reduce this angle to fit within the principal range.

Question 2 (c):

Find the Cartesian equations for the curves given by z+1i=2|z + 1 - i| = \sqrt{2} and z1+i=5|z - 1 + i| = \sqrt{5}, where z=x+iyz = x + iy with x,yRx, y \in \mathbb{R}.

  • First equation: z+1i=2|z + 1 - i| = \sqrt{2} means (x+1)2+(y1)2=2\sqrt{(x + 1)^2 + (y - 1)^2} = \sqrt{2}. Squaring both sides, we get: (x+1)2+(y1)2=2(x + 1)^2 + (y - 1)^2 = 2 This represents a circle with center (1,1)(-1, 1) and radius 2\sqrt{2}.

  • Second equation: z1+i=5|z - 1 + i| = \sqrt{5} means (x1)2+(y+1)2=5\sqrt{(x - 1)^2 + (y + 1)^2} = \sqrt{5}. Squaring both sides, we get: (x1)2+(y+1)2=5(x - 1)^2 + (y + 1)^2 = 5 This represents a circle with center (1,1)(1, -1) and radius 5\sqrt{5}.

Question 2 (d):

This part asks to find the x- and y-intercepts of the two curves derived in part (c).

  • For each circle equation, we substitute y=0y = 0 for the x-intercepts and x=0x = 0 for the y-intercepts and solve the corresponding equations.

Let me know if you'd like detailed calculations for any of these steps or clarifications!

Follow-up Questions:

  1. How do we convert a complex number to polar form in general?
  2. What is the geometric interpretation of the modulus and argument of a complex number?
  3. How do we find the conjugate of a complex number, and why is it useful?
  4. What is the significance of the principal argument, and how is it calculated?
  5. How can we derive Cartesian equations from complex expressions?

Tip:

When dealing with complex numbers, always remember that the modulus represents the "distance" from the origin, and the argument represents the "angle" from the positive real axis.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Coordinates
Modulus and Argument
Conjugates of Complex Numbers
Cartesian Equations

Formulas

Polar form: z = r * e^(iθ)
Modulus: |z| = sqrt(x^2 + y^2)
Argument: θ = atan(y/x)
Conjugate: z* = a - bi
Equation of a circle in Cartesian coordinates

Theorems

Euler's Formula: e^(iθ) = cos(θ) + i*sin(θ)
Properties of Modulus
Relationship between complex modulus and Cartesian plane

Suitable Grade Level

University Level - Calculus