CALCU-a. Find all solutions of w^4=-4 in polar for

a. Find all solutions of

w^{4}=-4

in polar form,

r e^{i \theta}

. State the modulus of the solutions (all solutions have the same modulus):

r=

List the principal argument of each solution:

\theta=

\square

Solution by Steps

step 1

To find the solutions in polar form, we first express the number -4 as a complex number in polar form

step 2

The complex number -4 can be written as

4e^{i\pi}

since its modulus is 4 and its argument is

\pi

, as it lies on the negative real axis

step 3

The equation

w^4 = 4e^{i\pi}

can be solved by taking the fourth root of both sides

step 4

The fourth roots of

4e^{i\pi}

are given by

4^{1/4}e^{i(\pi/4 + k\pi/2)}

for

k = 0, 1, 2, 3

step 5

The modulus of each solution is

r = 4^{1/4}

step 6

The principal arguments of the solutions are

\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4

Answer

r = \sqrt[4]{4}

\theta = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4

Key Concept

Finding complex roots in polar form

Explanation

To find the complex roots of a number in polar form, we take the nth root of the modulus and divide the argument by n, adding

k\cdot2\pi/n

for each of the n roots.

4_sm1 / assignment_3 / 3 Previous Problem Problem List Next Problem Assignment 3: Problem 3 (1 point) Classify the following function: -

f_{5}(a+b i): \mathbb{C} \rightarrow \mathbb{C}

defined by

f_{5}(a+b i)=(2+3 a)+i\left(b^{2}-1\right)

\square

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Solution by Steps

step 1

Identify the real and imaginary parts of the function

step 2

The real part of the function is

2+3a

step 3

The imaginary part of the function is

b^2-1

step 4

Since both parts are polynomials in

a

and

b

, the function is a polynomial function

Answer

The function

f_5(a+bi) = (2+3a) + i(b^2-1)

is a polynomial function.

Key Concept

Classifying Complex Functions

Explanation

A function of a complex variable is classified based on the form of its real and imaginary parts. If both parts are polynomials, the function is a polynomial function.

(1) (a) Use your solution from Problem 2 in WebWork to find all complex solutions of

(z-2)^{4}+4=0

. Express your final answers in Cartesian form.

Solution by Steps

step 1

To find the complex solutions of the equation

(z-2)^{4}+4=0

, we first solve for

z

in the equation

(z-2)^{4}=-4

step 2

We can rewrite

-4

as

-4e^{i\pi}

since

e^{i\pi} = -1

. Thus, the equation becomes

(z-2)^{4} = -4e^{i\pi}

step 3

Taking the fourth root of both sides, we get

z-2 = \sqrt[4]{-4e^{i\pi}}

. The fourth roots of

e^{i\pi}

are

e^{i(\pi/4 + k\pi/2)}

for

k = 0, 1, 2, 3

step 4

Therefore, the solutions for

z-2

are

e^{i\pi/4}

,

e^{i3\pi/4}

,

e^{i5\pi/4}

, and

e^{i7\pi/4}

step 5

Multiplying by the modulus

\sqrt[4]{4} = \sqrt{2}

, we get the solutions for

z-2

as

\sqrt{2}e^{i\pi/4}

,

\sqrt{2}e^{i3\pi/4}

,

\sqrt{2}e^{i5\pi/4}

, and

\sqrt{2}e^{i7\pi/4}

step 6

Converting these to Cartesian form and adding 2 to each to solve for

z

, we get

z = 1 + i

,

z = 1 - i

,

z = 3 - i

, and

z = 3 + i

Answer

z = 1 + i

,

z = 1 - i

,

z = 3 - i

,

z = 3 + i

Key Concept

Solving complex equations by converting to polar form and taking roots

Explanation

The equation

(z-2)^{4}+4=0

can be solved by expressing the complex number in polar form, taking the fourth root, and then converting back to Cartesian form. The solutions are found by adding 2 to the real part of each root.

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