Math Problem Statement

Express z equals negative 2 plus j in polar form and hence find z Superscript 6​, converting your answer into cartesian form.

Solution

We are given z=2+jz = -2 + j, and we need to express it in polar form and then find z6z^6, converting the result back into Cartesian form.

Step 1: Express zz in polar form

The polar form of a complex number z=x+jyz = x + jy is given by:

z=r(cosθ+jsinθ)z = r \left( \cos \theta + j \sin \theta \right)

where:

  • rr is the modulus of zz,
  • θ\theta is the argument (angle) of zz.

For z=2+jz = -2 + j:

  • The real part x=2x = -2,
  • The imaginary part y=1y = 1.

1.1 Modulus of zz (r)

The modulus rr is calculated as:

r=x2+y2=(2)2+12=4+1=5r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}

So, r=5r = \sqrt{5}.

1.2 Argument of zz (θ)

The argument θ\theta is calculated using:

θ=tan1(yx)=tan1(12)\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{1}{-2}\right)

Since the point (2,1)(-2, 1) lies in the second quadrant, we adjust the argument:

θ=πtan1(12)\theta = \pi - \tan^{-1}\left(\frac{1}{2}\right)

Using the approximation tan1(0.5)0.464\tan^{-1}(0.5) \approx 0.464 radians:

θπ0.464=2.677 radians\theta \approx \pi - 0.464 = 2.677 \text{ radians}

Thus, the polar form of zz is:

z=5(cos(2.677)+jsin(2.677))z = \sqrt{5} \left( \cos(2.677) + j \sin(2.677) \right)

Step 2: Find z6z^6 in polar form

In polar form, raising a complex number to a power is straightforward. If z=r(cosθ+jsinθ)z = r \left( \cos \theta + j \sin \theta \right), then:

zn=rn(cos(nθ)+jsin(nθ))z^n = r^n \left( \cos(n\theta) + j \sin(n\theta) \right)

For z6z^6:

  • r6=(5)6=53=125r^6 = (\sqrt{5})^6 = 5^3 = 125,
  • 6θ=6×2.677=16.062 radians6\theta = 6 \times 2.677 = 16.062 \text{ radians}.

We subtract multiples of 2π2\pi to bring 6θ6\theta within the range [0,2π][0, 2\pi]:

16.0622π×216.06212.566=3.496 radians.16.062 - 2\pi \times 2 \approx 16.062 - 12.566 = 3.496 \text{ radians}.

Thus, the polar form of z6z^6 is:

z6=125(cos(3.496)+jsin(3.496))z^6 = 125 \left( \cos(3.496) + j \sin(3.496) \right)

Step 3: Convert z6z^6 back to Cartesian form

We use the approximations cos(3.496)0.939\cos(3.496) \approx -0.939 and sin(3.496)0.343\sin(3.496) \approx -0.343:

z6=125(0.939+j(0.343))z^6 = 125 \left( -0.939 + j(-0.343) \right)

Multiplying through by 125:

z6=117.37542.875jz^6 = -117.375 - 42.875j

Thus, the Cartesian form of z6z^6 is:

z6=117.37542.875jz^6 = -117.375 - 42.875j

Final Answer:

z6=117.37542.875jz^6 = -117.375 - 42.875j

Do you need more details on any step or have any questions?
Here are 5 related questions:

  1. How do you convert a complex number from polar to Cartesian form?
  2. What is De Moivre's Theorem, and how is it used for powers of complex numbers?
  3. What is the geometric interpretation of the modulus and argument of a complex number?
  4. How can you find the argument of a complex number in different quadrants?
  5. How does multiplying complex numbers in polar form affect their modulus and argument?

Tip: Always ensure to adjust the argument based on the quadrant where the complex number lies.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Polar Coordinates
Powers of Complex Numbers

Formulas

Polar form: z = r(cosθ + j sinθ)
Modulus: r = sqrt(x^2 + y^2)
Argument: θ = tan^{-1}(y/x)
De Moivre's Theorem: z^n = r^n (cos(nθ) + j sin(nθ))

Theorems

De Moivre's Theorem

Suitable Grade Level

Grades 11-12