CALCU-This is basic algebra. Please answer all of

This is basic algebra. Please answer all of the questions.

Generated Graph

Solution by Steps

step 1

Expand the expression:

(5 + j3)(2 - j) - (3 + j)

step 2

Use distributive property:

5 \cdot 2 + 5 \cdot (-j) + j3 \cdot 2 + j3 \cdot (-j) - (3 + j)

step 3

Simplify:

10 - 5j + 6j - 3j^2 - 3 - j

step 4

Since

j^2 = -1

:

10 - 5j + 6j + 3 - 3 - j

step 5

Combine like terms:

10 + 3 - 5j + 6j - j - 3

step 6

Final result:

10 + 3 - 3 - j = 10 - j

(b)

(1-j 2)^{2}

step 1

Expand the expression:

(1 - j2)^2

step 2

Use distributive property:

(1 - j2)(1 - j2)

step 3

Simplify:

1 - j2 - j2 + (j2)^2

step 4

Since

j^2 = -1

:

1 - 2j - 2j + 4

step 5

Combine like terms:

1 + 4 - 4j

step 6

Final result:

5 - 4j

(c)

\frac{5-j 8}{3-j 4}

step 1

Simplify the expression:

\frac{5 - j8}{3 - j4}

step 2

Multiply numerator and denominator by the conjugate of the denominator:

\frac{(5 - j8)(3 + j4)}{(3 - j4)(3 + j4)}

step 3

Simplify:

\frac{15 + 20j - 24 - j32}{9 + 16}

step 4

Combine like terms:

\frac{-9 + 20j - 32j}{25}

step 5

Final result:

\frac{-9 - 12j}{25} = -\frac{9}{25} - \frac{12j}{25}

(d)

\frac{1}{2}(1+j)^{2}

step 1

Expand the expression:

\frac{1}{2}(1 + j)^2

step 2

Use distributive property:

\frac{1}{2}(1 + 2j + j^2)

step 3

Since

j^2 = -1

:

\frac{1}{2}(1 + 2j - 1)

step 4

Combine like terms:

\frac{1}{2}(2j)

step 5

Final result:

j

(e)

\frac{1}{5-j 3}-\frac{1}{5+j 3}

step 1

Simplify the expression:

\frac{1}{5 - j3} - \frac{1}{5 + j3}

step 2

Multiply numerator and denominator by the conjugate of the denominator:

\frac{(5 + j3) - (5 - j3)}{(5 - j3)(5 + j3)}

step 3

Simplify:

\frac{5 + j3 - 5 + j3}{25 + 9}

step 4

Combine like terms:

\frac{2j3}{34}

step 5

Final result:

\frac{6j}{34} = \frac{3j}{17}

Question 2 (a)

x^{2}+2 x+2=0

step 1

Use the quadratic formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

step 2

Substitute

a = 1

,

b = 2

,

c = 2

:

x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}

step 3

Simplify:

x = \frac{-2 \pm \sqrt{4 - 8}}{2}

step 4

Since

\sqrt{-4} = 2j

:

x = \frac{-2 \pm 2j}{2}

step 5

Final result:

x = -1 \pm j

(b)

x^{3}+8=0

step 1

Rewrite the equation:

x^3 = -8

step 2

Find the cube roots:

x = \sqrt[3]{-8}

step 3

Simplify:

x = -2

step 4

Find the complex roots:

x = 1 - j\sqrt{3}

and

x = 1 + j\sqrt{3}

step 5

Final result:

x = -2, 1 - j\sqrt{3}, 1 + j\sqrt{3}

Question 3 Find

z

such that

z z^{*}+3(z-z^{*})=13+j 12

step 1

Rewrite the equation:

z z^* + 3(z - z^*) = 13 + j12

step 2

Simplify:

z z^* + 3z - 3z^* = 13 + j12

step 3

Use the property

z^* = \overline{z}

:

z \overline{z} + 3z - 3\overline{z} = 13 + j12

step 4

Solve for

z

:

z = \pm \sqrt{12j + 13}

step 5

Final result:

z = \pm \sqrt{12j + 13}

Question 4 With

z=2-j 3

, find (a)

j z

step 1

Multiply

z

by

j

:

j(2 - j3)

step 2

Simplify:

2j - j^2 3

step 3

Since

j^2 = -1

:

2j + 3

step 4

Final result:

3 + 2j

(b)

z^*

step 1

Find the conjugate of

z

:

2 - j3

step 2

Conjugate:

2 + j3

step 3

Final result:

2 + j3

(c)

\frac{1}{z}

step 1

Find the reciprocal of

z

:

\frac{1}{2 - j3}

step 2

Multiply numerator and denominator by the conjugate of the denominator:

\frac{1 \cdot (2 + j3)}{(2 - j3)(2 + j3)}

step 3

Simplify:

\frac{2 + j3}{4 + 9}

step 4

Final result:

\frac{2 + j3}{13}

(d)

\left(z^{*}\right)^{*}

step 1

Find the conjugate of the conjugate of

z

:

(2 + j3)^*

step 2

Conjugate:

2 - j3

step 3

Final result:

2 - j3

Question 5 Find

z=\frac{z_{1}+z_{2} z_{3}}{z_{2}+z_{3}}

when

z_{1}=2+j 3, z_{2}=3+j 4

, and

z_{3}=-5+j 12

step 1

Substitute

z_1

,

z_2

, and

z_3

:

z = \frac{2 + j3 + (3 + j4)(-5 + j12)}{3 + j4 + (-5 + j12)}

step 2

Simplify the numerator:

2 + j3 + (3 \cdot -5 + 3 \cdot j12 + j4 \cdot -5 + j4 \cdot j12)

step 3

Simplify:

2 + j3 + (-15 + j36 - j20 - 48)

step 4

Combine like terms:

2 + j3 - 15 - 48 + j36 - j20

step 5

Final result:

-61 + j19

Question 6 Find the values of the real numbers

x

and

y

which satisfy the equation

\frac{2+x-j y}{3 x+j y}=1+j 2

step 1

Cross-multiply:

(2 + x - jy)(1 + j2) = (3x + jy)

step 2

Expand:

2 + x + 2j + j2x - yj - 2y = 3x + jy

step 3

Separate real and imaginary parts:

2 + x - 2y = 3x

and

2j + j2x - yj = jy

step 4

Solve for

x

and

y

:

2 - 2y = 2x

and

2 + 2x - y = y

step 5

Final result:

x = -1

and

y = -9/2

Question 7 Show in an Argand diagram the points representing the following complex numbers. Find the modulus and argument of each of the complex numbers given. (a)

1+j

step 1

Plot the point

(1, 1)

on the Argand diagram

step 2

Find the modulus:

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

step 3

Find the argument:

\tan^{-1}(1/1) = \pi/4

step 4

Final result: Modulus =

\sqrt{2}

, Argument =

\pi/4

(b)

\sqrt{3}-j

step 1

Plot the point

(\sqrt{3}, -1)

on the Argand diagram

step 2

Find the modulus:

\sqrt{(\sqrt{3})^2 + (-1)^2} = 2

step 3

Find the argument:

\tan^{-1}(-1/\sqrt{3}) = -\pi/6

step 4

Final result: Modulus =

2

, Argument =

-\pi/6

(c)

-1+j \sqrt{3}

step 1

Plot the point

(-1, \sqrt{3})

on the Argand diagram

step 2

Find the modulus:

\sqrt{(-1)^2 + (\sqrt{3})^2} = 2

step 3

Find the argument:

\tan^{-1}(\sqrt{3}/-1) = 2\pi/3

step 4

Final result: Modulus =

2

, Argument =

2\pi/3

(d)

-1-j \sqrt{3}

step 1

Plot the point

(-1, -\sqrt{3})

on the Argand diagram

step 2

Find the modulus:

\sqrt{(-1)^2 + (-\sqrt{3})^2} = 2

step 3

Find the argument:

\tan^{-1}(-\sqrt{3}/-1) = -2\pi/3

step 4

Final result: Modulus =

2

, Argument =

-2\pi/3

Question 8 Express the complex numbers in polar form and exponential form. (a)

j

step 1

Polar form:

1e^{j\pi/2}

step 2

Exponential form:

e^{j\pi/2}

step 3

Final result:

e^{j\pi/2}

(b)

1-j

step 1

Polar form:

\sqrt{2}e^{-j\pi/4}

step 2

Exponential form:

\sqrt{2}e^{-j\pi/4}

step 3

Final result:

\sqrt{2}e^{-j\pi/4}

(c)

\sqrt{3}-j \sqrt{3}

step 1

Polar form:

2e^{-j\pi/6}

step 2

Exponential form:

2e^{-j\pi/6}

step 3

Final result:

2e^{-j\pi/6}

(d)

-2+j

step 1

Polar form:

\sqrt{5}e^{j\pi/2}

step 2

Exponential form:

\sqrt{5}e^{j\pi/2}

step 3

Final result:

\sqrt{5}e^{j\pi/2}

(e)

(2-j)(2+j)

step 1

Simplify:

4 - j^2

step 2

Since

j^2 = -1

:

4 + 1 = 5

step 3

Final result:

5

(f)

7-j 5

step 1

Polar form:

\sqrt{74}e^{-j\pi/4}

step 2

Exponential form:

\sqrt{74}e^{-j\pi/4}

step 3

Final result:

\sqrt{74}e^{-j\pi/4}

Question 9 Express

z=\frac{(2-j)(3+j 2)}{(3-j 4)}

in the form

x+j y

and also in polar form.

step 1

Simplify the expression:

\frac{(2 - j)(3 + j2)}{3 - j4}

step 2

Multiply numerator and denominator by the conjugate of the denominator:

\frac{(2 - j)(3 + j2)(3 + j4)}{(3 - j4)(3 + j4)}

step 3

Simplify:

\frac{6 + j8 - j3 - j^2 2}{9 + 16}

step 4

Combine like terms:

\frac{6 - 2 + j8 - j3}{25}

step 5

Final result:

\frac{4 + j5}{25} = \frac{4}{25} + \frac{5j}{25}

Question 10 Given

z_{1}=2 e^{J \frac{\pi}{3}}

and

z_{2}=4 e^{-\frac{2 j \pi}{3}}

, find the modulus and argument of (a)

z_{1}^{3} z_{2}^{2}

step 1

Find the modulus:

|z_1^3 z_2^2| = |2^3 \cdot 4^2| = 64

step 2

Find the argument:

\arg(z_1^3 z_2^2) = 3 \cdot \frac{\pi}{3} + 2 \cdot -\frac{2\pi}{3} = \pi - \frac{4\pi}{3} = -\frac{\pi}{3}

step 3

Final result: Modulus =

64

, Argument =

-\frac{\pi}{3}

(b)

z_{1}^{2} z_{2}^{4}

step 1

Find the modulus:

|z_1^2 z_2^4| = |2^2 \cdot 4^4| = 1024

step 2

Find the argument:

\arg(z_1^2 z_2^4) = 2 \cdot \frac{\pi}{3} + 4 \cdot -\frac{2\pi}{3} = \frac{2\pi}{3} - \frac{8\pi}{3} = -2\pi

step 3

Final result: Modulus = $

This is basic algebra. Please answer all the questions.

Generated Graph

Solution by Steps

step 1

Given the equation

\operatorname{Re}\left(\frac{z+j}{z-j}\right)=1

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: - \operatorname{Re}(z)<\operatorname{Re}(j) and

\left(\operatorname{Re}(j) (\operatorname{Re}(z) - \operatorname{Re}(j))\right)/\operatorname{Im}(j) + \operatorname{Im}(z) = \operatorname{Im}(j)

and \operatorname{Im}(j)<0 - \operatorname{Re}(z)>\operatorname{Re}(j) and

\left(\operatorname{Re}(j) (\operatorname{Re}(z) - \operatorname{Re}(j))\right)/\operatorname{Im}(j) + \operatorname{Im}(z) = \operatorname{Im}(j)

and \operatorname{Im}(j)<0 -

\operatorname{Re}(j) = \operatorname{Re}(z)

and \operatorname{Im}(z)<0 and

\operatorname{Im}(j) = 0

and \operatorname{Re}(j)<0 -

\operatorname{Re}(j) = \operatorname{Re}(z)

and \operatorname{Im}(z)>0 and

\operatorname{Im}(j) = 0

and \operatorname{Re}(j)<0

# (b)

\left|\frac{z+j}{z-j}\right|=3

step 1

Given the equation

\left|\frac{z+j}{z-j}\right|=3

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: -

z = \frac{j (-1 + 3 e^{i n})}{1 + 3 e^{i n}}

and

j \neq 0

and

n \in \mathbb{R}

-

z = 2j

and j<0 -

z = \frac{j}{2}

and j<0 -

z = \frac{j}{2}

and j>0

# (c)

\tan \arg \left(\frac{z+j}{z-j}\right)=\sqrt{3}

step 1

Given the equation

\tan \arg \left(\frac{z+j}{z-j}\right)=\sqrt{3}

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, there are no real solutions

Question 16 # (a)

|z-1|=2

step 1

Given the equation

|z-1|=2

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: -

z = 1 + 2 e^{i n}

and

n \in \mathbb{R}

-

z = -1

-

z = 3

# (b)

|z-2-j 3|=4

step 1

Given the equation

|z-2-j 3|=4

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: -

z = 3j + 4 e^{i n} + 2

and

n \in \mathbb{R}

-

z = 3(j + 2)

-

z = 3j - 2

# (c)

|z-4|=3|z+1|

step 1

Given the equation

|z-4|=3|z+1|

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: -

-\frac{7}{2} \leq \operatorname{Re}(z) \leq \frac{1}{4}

and

\operatorname{Im}(z) = \pm \left(\frac{1}{2} \sqrt{-4 \operatorname{Re}(z)^2 - 13 \operatorname{Re}(z) + \frac{7}{2}}\right)

-

z = -\frac{7}{2}

-

z = \frac{1}{4}

Question 17 #

\left|\frac{z+1}{z-2}\right|=2

step 1

Given the equation

\left|\frac{z+1}{z-2}\right|=2

, we need to find the locus of

z

in the complex plane

step 2

From the Asksia-LL calculator, the solutions are: -

z = \frac{1 + 4 e^{i n}}{-1 + 2 e^{i n}}

and

n \in \mathbb{R}

-

z = 1

-

z = 5

Question 18 # The circle

x^{2}+y^{2}+4 x=0

is taken to lie on the Argand diagram. Describe the circle in terms of

z

.

step 1

Given the equation

x^{2}+y^{2}+4 x=0

, we need to describe the circle in terms of

z

step 2

The equation can be rewritten as

(x+2)^2 + y^2 = 4

step 3

In terms of

z

, this represents a circle with center

-2

and radius

2

Answer

The loci and Cartesian equations for the given complex number problems are provided above.

Key Concept

Complex number loci and Cartesian equations

Explanation

The solutions involve finding the loci of complex numbers in the Argand diagram and converting complex number equations to Cartesian form.

This is basic algebra. Please answer all the questions.

Generated Graph

Solution by Steps

step 1

Given

z_{1} = 2 e^{j \frac{\pi}{3}}

and

z_{2} = 4 e^{-j \frac{2\pi}{3}}

, we need to find the modulus and argument of

z_{1}^{3} z_{2}^{2}

step 2

Calculate the modulus:

|z_{1}^{3} z_{2}^{2}| = |2 e^{j \frac{\pi}{3}}|^3 \cdot |4 e^{-j \frac{2\pi}{3}}|^2 = 2^3 \cdot 4^2 = 8 \cdot 16 = 128

step 3

Calculate the argument:

\arg(z_{1}^{3} z_{2}^{2}) = 3 \cdot \frac{\pi}{3} + 2 \cdot \left(-\frac{2\pi}{3}\right) = \pi - \frac{4\pi}{3} = -\frac{\pi}{3}

Answer

Modulus: 128, Argument:

-\frac{\pi}{3}

Key Concept

Modulus and Argument of Complex Numbers

Explanation

The modulus is the product of the magnitudes, and the argument is the sum of the angles.

# (b)

z_{1}^{2} z_{2}^{4}

step 1

Given

z_{1} = 2 e^{j \frac{\pi}{3}}

and

z_{2} = 4 e^{-j \frac{2\pi}{3}}

, we need to find the modulus and argument of

z_{1}^{2} z_{2}^{4}

step 2

Calculate the modulus:

|z_{1}^{2} z_{2}^{4}| = |2 e^{j \frac{\pi}{3}}|^2 \cdot |4 e^{-j \frac{2\pi}{3}}|^4 = 2^2 \cdot 4^4 = 4 \cdot 256 = 1024

step 3

Calculate the argument:

\arg(z_{1}^{2} z_{2}^{4}) = 2 \cdot \frac{\pi}{3} + 4 \cdot \left(-\frac{2\pi}{3}\right) = \frac{2\pi}{3} - \frac{8\pi}{3} = -2\pi

Answer

Modulus: 1024, Argument:

-2\pi

Key Concept

Modulus and Argument of Complex Numbers

Explanation

The modulus is the product of the magnitudes, and the argument is the sum of the angles.

# (c)

\frac{z_{1}^{2}}{z_{2}^{3}}

step 1

Given

z_{1} = 2 e^{j \frac{\pi}{3}}

and

z_{2} = 4 e^{-j \frac{2\pi}{3}}

, we need to find the modulus and argument of

\frac{z_{1}^{2}}{z_{2}^{3}}

step 2

Calculate the modulus:

|\frac{z_{1}^{2}}{z_{2}^{3}}| = \frac{|2 e^{j \frac{\pi}{3}}|^2}{|4 e^{-j \frac{2\pi}{3}}|^3} = \frac{2^2}{4^3} = \frac{4}{64} = \frac{1}{16}

step 3

Calculate the argument:

\arg\left(\frac{z_{1}^{2}}{z_{2}^{3}}\right) = 2 \cdot \frac{\pi}{3} - 3 \cdot \left(-\frac{2\pi}{3}\right) = \frac{2\pi}{3} + 2\pi = \frac{8\pi}{3}

Answer

Modulus:

\frac{1}{16}

, Argument:

\frac{8\pi}{3}

Key Concept

Modulus and Argument of Complex Numbers

Explanation

The modulus is the quotient of the magnitudes, and the argument is the difference of the angles.

Question 11 # (a)

(1+j)^3

step 1

Use de Moivre's theorem to calculate

(1+j)^3

. First, convert

1+j

to polar form:

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

and

\theta = \tan^{-1}(1) = \frac{\pi}{4}

step 2

Apply de Moivre's theorem:

(\sqrt{2} e^{j \frac{\pi}{4}})^3 = (\sqrt{2})^3 e^{j \frac{3\pi}{4}} = 2\sqrt{2} e^{j \frac{3\pi}{4}}

Answer

2\sqrt{2} e^{j \frac{3\pi}{4}}

Key Concept

De Moivre's Theorem

Explanation

De Moivre's theorem allows us to raise complex numbers in polar form to a power by raising the modulus to the power and multiplying the argument by the power.

# (b)

(\sqrt{3}-j)^3

step 1

Use de Moivre's theorem to calculate

(\sqrt{3}-j)^3

. First, convert

\sqrt{3}-j

to polar form:

r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2

and

\theta = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}

step 2

Apply de Moivre's theorem:

(2 e^{-j \frac{\pi}{6}})^3 = 2^3 e^{-j \frac{3\pi}{6}} = 8 e^{-j \frac{\pi}{2}}

Answer

8 e^{-j \frac{\pi}{2}}

Key Concept

De Moivre's Theorem

Explanation

De Moivre's theorem allows us to raise complex numbers in polar form to a power by raising the modulus to the power and multiplying the argument by the power.

# (c)

(-1+j\sqrt{3})^3

step 1

Use de Moivre's theorem to calculate

(-1+j\sqrt{3})^3. First, convert

-1+j\sqrt{3}

to polar form:

r = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2

and

\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3}$

step 2

Apply de Moivre's theorem:

(2 e^{j \frac{2\pi}{3}})^3 = 2^3 e^{j 2\pi} = 8 e^{j 2\pi} = 8

Answer

8

Key Concept

De Moivre's Theorem

Explanation

De Moivre's theorem allows us to raise complex numbers in polar form to a power by raising the modulus to the power and multiplying the argument by the power.

# (d)

(-1-j\sqrt{3})^3

step 1

Use de Moivre's theorem to calculate

(-1-j\sqrt{3})^3

. First, convert

-1-j\sqrt{3}

to polar form:

r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = 2

and

\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = -\frac{2\pi}{3}

step 2

Apply de Moivre's theorem:

(2 e^{-j \frac{2\pi}{3}})^3 = 2^3 e^{-j 2\pi} = 8 e^{-j 2\pi} = 8

Answer

8

Key Concept

De Moivre's Theorem

Explanation

De Moivre's theorem allows us to raise complex numbers in polar form to a power by raising the modulus to the power and multiplying the argument by the power.

Question 12

step 1

Find the three values of

(8+j8)^{\frac{1}{3}}

. First, convert

8+j8

to polar form:

r = \sqrt{8^2 + 8^2} = 8\sqrt{2}

and

\theta = \tan^{-1}(1) = \frac{\pi}{4}

step 2

Apply the cube root:

(8\sqrt{2} e^{j \frac{\pi}{4}})^{\frac{1}{3}} = (8\sqrt{2})^{\frac{1}{3}} e^{j \frac{\pi}{12}} = 2\sqrt[3]{2} e^{j \frac{\pi}{12}}

step 3

Find the three roots:

2\sqrt[3]{2} e^{j \frac{\pi}{12}}, 2\sqrt[3]{2} e^{j \frac{\pi}{12} + \frac{2\pi}{3}}, 2\sqrt[3]{2} e^{j \frac{\pi}{12} + \frac{4\pi}{3}}

Answer

2\sqrt[3]{2} e^{j \frac{\pi}{12}}, 2\sqrt[3]{2} e^{j \frac{\pi}{12} + \frac{2\pi}{3}}, 2\sqrt[3]{2} e^{j \frac{\pi}{12} + \frac{4\pi}{3}}

Key Concept

Cube Roots of Complex Numbers

Explanation

The cube roots of a complex number are found by taking the cube root of the modulus and dividing the argument by 3, then adding multiples of

\frac{2\pi}{3}

.

Question 13 # (a)

(\sqrt{3}-j)^{1/4}

step 1

Find the fourth root of

(\sqrt{3}-j)

. First, convert

\sqrt{3}-j

to polar form:

r = 2

and

\theta = -\frac{\pi}{6}

step 2

Apply the fourth root:

(2 e^{-j \frac{\pi}{6}})^{\frac{1}{4}} = 2^{\frac{1}{4}} e^{-j \frac{\pi}{24}}

Answer

2^{\frac{1}{4}} e^{-j \frac{\pi}{24}}

Key Concept

Roots of Complex Numbers

Explanation

The roots of a complex number are found by taking the root of the modulus and dividing the argument by the root, then adding multiples of

\frac{2\pi}{n}

.

# (b)

(j8)^{1/3}

step 1

Find the cube root of

(j8)

. First, convert

j8

to polar form:

r = 8

and

\theta = \frac{\pi}{2}

step 2

Apply the cube root:

(8 e^{j \frac{\pi}{2}})^{\frac{1}{3}} = 8^{\frac{1}{3}} e^{j \frac{\pi}{6}} = 2 e^{j \frac{\pi}{6}}

Answer

2 e^{j \frac{\pi}{6}}

Key Concept

Roots of Complex Numbers

Explanation

The roots of a complex number are found by taking the root of the modulus and dividing the argument by the root, then adding multiples of

\frac{2\pi}{n}

.

# (c)

(-1)^{1/4}

step 1

Find the fourth root of

(-1)

. First, convert

-1

to polar form:

r = 1

and

\theta = \pi

step 2

Apply the fourth root:

(1 e^{j \pi})^{\frac{1}{4}} = 1^{\frac{1}{4}} e^{j \frac{\pi}{4}} = e^{j \frac{\pi}{4}}

Answer

e^{j \frac{\pi}{4}}

Key Concept

Roots of Complex Numbers

Explanation

The roots of a complex number are found by taking the root of the modulus and dividing the argument by the root, then adding multiples of

\frac{2\pi}{n}

.

Question 14 # (a)

\operatorname{Re} z=5

step 1

The locus of

z

when

\operatorname{Re} z = 5

is a vertical line in the complex plane where the real part of

z

is always 5

Answer

Vertical line at

\operatorname{Re} z = 5

Key Concept

Locus of Complex Numbers

Explanation

The locus of a complex number with a fixed real part is a vertical line at that real part.

# (b)

|z-1|=3

step 1

The locus of

z

when

|z-1|=3

is a circle in the complex plane with center at

1

and radius

3

Answer

Circle with center at

1

and radius

3

Key Concept

Locus of Complex Numbers

Explanation

The locus of a complex number with a fixed distance from a point is a circle centered at that point with the given radius.

# (c)

\left|\frac{z-1}{z+1}\right|=3

step 1

The locus of

z

when

\left|\frac{z-1}{z+1}\right|=3

is a circle in the complex plane. This represents a circle with a specific transformation

Answer

Circle with a specific transformation

Key Concept

Locus of Complex Numbers

Explanation

The locus of a complex number with a fixed ratio of distances from two points is a circle.

# (d)

\arg (z \mid-2)=\pi / 4

step 1

The locus of

z

when

\arg (z \mid-2)=\pi / 4

is a ray in the complex plane starting from

-2

and making an angle of

\pi/4

with the positive real axis

Answer

Ray starting from

-2

making an angle of

\pi/4

with the positive real axis

Key Concept

Locus of Complex Numbers

Explanation

The locus of a complex number with a fixed argument relative to a point is a ray starting from that point at the given angle.

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