Exercise 1 - Standard Form

Write these Complex numbers in Standard Form

a. \frac{2}{\sqrt{-2}}

b. \sqrt{-27}\cdot\sqrt{-3}

c.\sqrt{-32}-\sqrt{16}-\sqrt{-8}+\sqrt{4}

d. (3-2i)i

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Exercise 2 - Addition and Subtraction and the Complex Plane

Perform the addition or subtraction and draw the new Complex number

a. z_1+z_2=(3-4i)+(-3+4i)

b. z_1+z_2=(2-5i)i+(1+4i)i

c. z_1-z_2+z_3-z_4=(2+i)-(3-2i)+(1-5i)-(4+3i)

d. z_1-z_2=(4-6i)-(-5-3i)

Exercise 3 - Multiplication, Modulus and the Complex Plane

Perform the multiplication, draw the new Complex number and find the modulus

a. (3+2i)(-3-2i)

b. (3+4i)(1+i)

c. (-1-\sqrt{3}i)(1-\sqrt{3}i)

d. (2+2i)(3+3i)(1-i)

e. (3+i)(2+i)(1+2i)(1+3i)

Exercise 4 - Powers of (1+i) and the Complex Plane

Perform the following calculations on z=1+i and state the position in the plane, how much you rotated and the modulus at each step

a. (1+i)^{2}

b. (1+i)^{3}

c. (1+i)^{4}

d. (1+i)^{5}

e. (1+i)^{6}

f. (1+i)^{7}

g. (1+i)^{8}

Exercise 5 - Opposites, Conjugates and Inverses

Find -z, z^{*} and z^{-1} (if they exist) for each of the following

a. z=4+4i

b. z=3-3i

c. z=0+2i

d. z=3+4i

e. z=-5-12i

f. z=2+0i

Exercise 6 - Reference Angles

Find the fractions that make the equations true

a. What is \frac{a}{b} in the equation \frac{a\pi}{b}-2\pi=\frac{5\pi}{6}?

b. What is \frac{a}{b} in the equation \frac{a\pi}{b}-2\pi=\frac{5\pi}{3}?

c. What is \frac{a}{b} in the equation \frac{129\pi}{6}-\frac{a\pi}{b}=\frac{3\pi}{2}?

Exercise 7- Division

Perform the division

a. \frac{z_1}{z_2}=\frac{\sqrt{2}+\sqrt{-3}}{\sqrt{2}-\sqrt{-3}}

b. \frac{z_1}{z_2}=\frac{5-i}{i}

c. \frac{z_1}{z_2}=\frac{-4i}{i-3}

d. \frac{z_1}{z_2}=\frac{2-3i}{2-i}

e. \frac{z_1}{z_2}=\frac{(2-3i)-(3+2i)}{(3+2i)-(2+i)}

Exercise 8 - Special Triangles and Arguments

Use special triangles to find a Complex number that has each of these arguments

a. \theta=\frac{3\pi}{4}

b. \theta=\frac{3\pi}{2}

c. \theta=\frac{\pi}{6}

d. \theta=\frac{5\pi}{3}

e. \theta=\frac{7\pi}{6}

f. \theta=-\frac{3\pi}{4}

g. \theta=-\frac{2\pi}{3}

Exercise 9 - Polar Form of Complex Numbers

Determine the Polar Form for each of these Complex numbers

a. z=\sqrt{3}+i

b. z=3-3i

c. z=1-\sqrt{3}i

d. z=-2i

e. z=-2-2i

f. z=-\sqrt{3}+i

g. z=4

Exercise 10 - Roots of Equations

Solve for the roots of these equations

a. (x-3)^{2}=-16

b. (x+iy)^{2}=5-12i

c. z^{2}=-2i

Exercise 11 - Powers of a Complex Number

Calculate the following numbers

a. (1-\sqrt{3}i)^{3}

b. \frac{(1+i)^{3}}{(1-i)^{5}}

Expand (\cos{\theta}+i\sin{\theta})^{6} to find

c. \cos{6\theta}

d. \sin{6\theta}

If \theta=\frac{\pi}{4} find the explicit answer for

e. (\cos{\theta}+i\sin{\theta})^{3}

f. (\cos{\theta}+i\sin{\theta})^{4}

Exercise 12 - Complex Roots

a. Calculate the 4th roots of unity z^{4}=1

b. Calculate the 6th roots of unity z^{6}=1

c. Calculate (3+4i)^{\frac{1}{2}}

d. Calculate (-1+\sqrt{3})^{\frac{1}{2}}

e. Use the Complex version of the Quadratic Formula to obtain the roots to the equation z^{2}-(3+2i)z+(1+3i)=0

Solutions for Exercises 1-12

Solutions for Exercise 1 - Standard Form

Write these Complex numbers in Standard Form

a. \frac{2}{\sqrt{-2}}

\frac{2}{\sqrt{-2}}=\frac{2}{\sqrt{2}i} then

\frac{2}{\sqrt{2}i}(\frac{\sqrt{2}i}{\sqrt{2}i})=\frac{2\sqrt{2}i}{-2}=-\sqrt{2}i

b. \sqrt{-27}\cdot\sqrt{-3}

\sqrt{-27}\cdot\sqrt{-3}=\sqrt{27}i\cdot\sqrt{3}i=\sqrt{81}i^{2}=-9

c.\sqrt{-32}-\sqrt{16}-\sqrt{-8}+\sqrt{4}

\sqrt{-32}-\sqrt{16}-\sqrt{-8}+\sqrt{4}=4\sqrt{2}i-4-2\sqrt{2}i+2=(-4+2)+(4\sqrt{2}-2\sqrt{2})i=-2+2\sqrt{2}i

d. (3-2i)i

(3-2i)i=3i-2i^{2}=2+3i

Solutions for Exercise 2 - Addition and Subtraction and the Complex Plane

Perform the addition or subtraction and draw the new Complex number

a. z_1+z_2=(3-4i)+(-3+4i)

(3-4i)+(-3+4i)=(3-3)+(-4+4)i=0+0i

Addition Answer a

b. z_1+z_2=(2-5i)i+(1+4i)i

(2-5i)i+(1-4i)i=(2i-5i^{2})+(i-4i^{2})=(5+2i)+(4+i)=(5+4)+(2+1)i=9+3i

Addition Exercise b

c. z_1-z_2+z_3-z_4=(2+i)-(3-2i)+(1-5i)-(4+3i)

(2+i)-(3-2i)+(1-5i)-(4+3i)=(2-3+1-4)+(1+2-5-3)i=-4-5i

Addition Exercise c

d. z_1-z_2=(4-6i)-(-5-3i)

(4-6i)-(-5-3i)=(4-(-5))+(-6-(-3))i=9-3i

Addition Exercise d

Solutions for Exercise 3 - Multiplication, Modulus and the Complex Plane

Perform the multiplication, draw the new Complex number and find the modulus

a. z=(3+2i)(-3-2i)

z=(3+2i)(-3-2i)=(-9+4)+(-6-6)i=-5-12i

modulus |z|=\sqrt{(-5)^{2}+(-12)^{2}}=\sqrt{25+144}=\sqrt{169}=13

Complex Multiplication Exercise a

b. z=(3+4i)(1+i)

z=(3+4i)(1+i)=(3-4)+(3+4)i=-1+7i

modulus |z|=\sqrt{(-1)^{2}+(7)^{2}}=\sqrt{1+49}=\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}

Complex Multiplication Exercise b

c. z=(-1-\sqrt{3}i)(1-\sqrt{3}i)

z=(-1-\sqrt{3}i)(1-\sqrt{3}i)=(-1-3)+(\sqrt{3}-\sqrt{3})i=-4+0i

modulus |z|=\sqrt{(-4)^{2}+(0)^{2}}=4

Complex Multiplication Exercise c

d. z=(2+2i)(3+3i)(1-i)

z=(2+2i)(3+3i)(1-i)=((6-6)+(6+6)i)(1-i)=(0+12i)(1-i)=(0+12)+(0+12)i=12+12i

modulus |z|=\sqrt{(12)^{2}+(12)^{2}}=\sqrt{144+144}=\sqrt{288}=\sqrt{144\times2}=12\sqrt{2}

Complex Multiplication Exercise d

e. z=(3+i)(2+i)(1+2i)(1+3i)

z=(3+i)(2+i)(1+2i)(1+3i)=((6-1)+(3+2)i)((1-6)+(3+2)i)=(5+5i)(-5+5i)=(-25-25)+(25-25)i=-50+0i

modulus |z|=\sqrt{(-50)^{2}}=50

Solutions for Exercise 4 - Powers of (1+i) and the Complex Plane

Perform the following calculations on z=1+i and state the position in the plane, how much you rotated and the modulus at each step

Powers of (1+i)

a. (1+i)^{2}

(1+i)(1+i)=(1-1)+(1+1)i=0+2i

rotation and position: \frac{\pi}{4} rotation to the positive y-axis at point (0,2)

all of the rotations will be \frac{\pi}{4} radians

modulus: |(1+i)^{2}|=|1+i||1+i|=(\sqrt{2})(\sqrt{2})=2

b. (1+i)^{3}

(1+i)^{3}=((1+i)^{2})(1+i)=(0+2i)(1+i)=(0-2)+(0+2)i=-2+2i

location: halfway between the positive y-axis and negative x-axis at the point (-2,2)

angle: \frac{\pi}{4}

modulus: |(1+i)^{3}|=|(1+i)^{2}||1+i|=2\sqrt{2}

c. (1+i)^{4}

(1+i)^{4}=((1+i)^{3})(1+i)=(-2+2i)(1+i)=(-2-2)+(-2+2)i=-4+0i

location: on the negative x-axis at the point (-4,0)

modulus: |(1+i)^{4}|=|(1+i)^{3}||1+i|=(2\sqrt{2})(\sqrt{2})=4

d. (1+i)^{5}

(1+i)^{5}=(-4+0i)(1+i)=(-4-0)+(-4-0)=-4-4i

location: halfway between the negative x-axis and the negative y-axis at the point (-4,-4)

modulus: |(1+i)^{5}|=|(1+i)^{4}||1+i|=(4)(\sqrt{2})=4\sqrt{2}

e. (1+i)^{6}

(1+i)^{6}=(-4-4i)(1+i)=(-4+4)+(-4-4)i=0-8i

location: on the negative y-axis at the point (0,-8)

modulus: |(1+i)^{6}|=|(1+i)^{5}||1+i|=(4\sqrt{2})(\sqrt{2})=8

f. (1+i)^{7}

(1+i)^{7}=(0-8i)(1+i)=(0+8)+(0-8)i=8-8i

location: halfway between the negative y-axis and the positive x-axis at the point (8,-8)

modulus: |(1+i)^{7}|=|(1+i)^{6}||1+i|=(8)(\sqrt{2})=8\sqrt{2}

g. (1+i)^{8}

(1+i)^{7}=(8-8i)(1+i)=(8+8)+(8-8)i=16+0i

location: the positive x-axis at the point (16,0)

modulus: |(1+i)^{8}|=|(1+i)^{7}||1+i|=(8\sqrt{2})(\sqrt{2})=16

Solutions for Exercise 5 - Opposites, Conjugates and Inverses

Find -z, z^{*} and z^{-1} (if they exist) for each of the following

a. z=4+4i

-z=-(4+4i)=-4-4i

z^{*}=4-4i

z^{-1}=\frac{1}{4+4i}=\frac{1}{4+4i}(\frac{4-4i}{4-4i})=\frac{4-4i}{32}=\frac{1-i}{8}

z^{-1}=\frac{z^{*}}{x^{2}+y^{2}}

b. z=3-3i

-z=-3+3i

z^{*}=3+3i

z^{-1}=\frac{3+3i}{18}=\frac{1+i}{6}

c. z=0+2i

-z=0-2i

z^{*}=0-2i

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

d. z=3+4i

-z=-3-4i

z^{*}=3-4i

z^{-1}=\frac{3-4i}{25}

e. z=-5-12i

-z=5+12i

z^{*}=-5+12i

z^{-1}=\frac{-5+12i}{169}

f. z=2+0i

-z=-2+0i

z^{*}=2-0i

z^{-1}=\frac{2}{4}=\frac{1}{2}

Solutions for Exercise 6 - Reference Angles

Find the fractions that make the equations true

a. What is \frac{a}{b} in the equation \frac{a\pi}{b}-2\pi=\frac{5\pi}{6}?

\frac{a\pi}{b}-2\pi=\frac{5\pi}{6}\implies\frac{a\pi}{b}=\frac{5\pi}{6}+\frac{12\pi}{6}=\frac{17\pi}{6}

b. What is \frac{a}{b} in the equation \frac{a\pi}{b}-2\pi=\frac{5\pi}{3}?

\frac{a\pi}{b}-2\pi=\frac{5\pi}{3}\implies\frac{a\pi}{b}=\frac{5\pi}{3}+\frac{6\pi}{3}=\frac{11\pi}{3}

c. What is \frac{a}{b} in the equation \frac{129\pi}{6}-\frac{a\pi}{b}=\frac{3\pi}{2}?

\frac{129\pi}{6}-\frac{a\pi}{b}=\frac{3\pi}{2}\implies\frac{a\pi}{b}=\frac{129\pi}{6}-\frac{9\pi}{6}=\frac{120\pi}{6}

Solutions for Exercise 7 - Division

Perform the division

a. \frac{z_1}{z_2}=\frac{\sqrt{2}+\sqrt{-3}}{\sqrt{2}-\sqrt{-3}}

\frac{z_1}{z_2}=\frac{\sqrt{2}+\sqrt{-3}}{\sqrt{2}-\sqrt{-3}}=

\frac{\sqrt{2}+\sqrt{3}i}{\sqrt{2}-\sqrt{3}i}(\frac{\sqrt{2}+\sqrt{3}i}{\sqrt{2}+\sqrt{3}i})=

\frac{(2-3)+(\sqrt{6}+\sqrt{6})i}{2+3}=\frac{-1+2\sqrt{6}}{5}

b. \frac{z_1}{z_2}=\frac{5-i}{i}

\frac{z_1}{z_2}=\frac{5-i}{i}(\frac{i}{i})=\frac{1+5i}{-1}=-1-5i

c. \frac{z_1}{z_2}=\frac{-4i}{-3+i}

\frac{z_1}{z_2}=\frac{-4i}{-3+i}(\frac{-3-i}{-3-i})=\frac{-4+12i}{10}=\frac{-2+6i}{5}

d. \frac{z_1}{z_2}=\frac{2-3i}{2-i}

\frac{z_1}{z_2}=\frac{2-3i}{2-i}(\frac{2+i}{2+i})=\frac{(4+3)+(2-6)i}{5}=\frac{7-4i}{5}

e. \frac{z_1}{z_2}=\frac{(2-3i)-(3+2i)}{(3+2i)-(2+i)}

\frac{z_1}{z_2}=\frac{(2-3i)-(3+2i)}{(3+2i)-(2+i)}=\frac{(2-3)+(-3-2)i}{(3-2)+(2-1)i}=

\frac{-1-5i}{1+i}(\frac{1-i}{1-i})=\frac{(-1-5)+(1-5)i}{2}=\frac{-6-4i}{2}=-3-2i

Solutions for Exercise 8 - Special Triangles and Arguments

Use special triangles to find a Complex number that has each of these arguments

a. \theta=\frac{3\pi}{4}

Quadrant II with negative and y positive

a 1:1:\sqrt{2} triangle for odd multiples of \frac{\pi}{4}

\cos{\frac{3\pi}{4}}=\frac{-\sqrt{2}}{2} and \sin{\frac{3\pi}{4}}=\frac{\sqrt{2}}{2}

z=-1+i, z=-2+2i etc. any number with |x|=|y| x<0 and y>0

b. \theta=\frac{3\pi}{2}

Negative y-axis with negative

z=0-i=-i, z=0-3i etc. any number with x=0 and y<0 

c. \theta=\frac{\pi}{6}

Quadrant I with and y both positive

a 1:\sqrt{3}:2 triangle with \cos{\frac{\pi}{6}}=\frac{\sqrt{3}}{2} and \sin{\frac{\pi}{6}}=\frac{1}{2}

z=\sqrt{3}+i, z=2\sqrt{3}+2i etc.

d. \theta=\frac{5\pi}{3}

Quadrant IV with x>0 and y<0

a 1:\sqrt{3}:2 triangle with \cos{\frac{5\pi}{3}}=\frac{1}{2} and \sin{\frac{5\pi}{3}}=-\frac{\sqrt{3}}{2}

z=1-\sqrt{3}i, z=3-3\sqrt{3}i etc.

e. \theta=\frac{7\pi}{6}

Quadrant III with both x<0 and y<0

a 1:\sqrt{3}:2 triangle with \cos{\frac{7\pi}{6}}=-\frac{\sqrt{3}}{2} and \sin{\frac{7\pi}{6}}=-\frac{1}{2}

z=-\sqrt{3}-i, z=-4\sqrt{3}-4i etc.

f. \theta=-\frac{3\pi}{4}

Quadrant III equivalent to angle \frac{5\pi}{4} with both x<0 and y<0

a 1:1:\sqrt{2} triangle with \cos{-\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}} and \sin{-\frac{3\pi}{4}}=-\frac{\sqrt{2}}{2}

z=-1-i, z=-3-3i etc.

g. \theta=-\frac{2\pi}{3}

Quadrant III equivalent to angle \frac{4\pi}{3} with x<0 and y<0

a 1:\sqrt{3}:2 triangle with \cos{-\frac{2\pi}{3}}=-\frac{1}{2} and \sin{-\frac{2\pi}{3}}=-\frac{\sqrt{3}}{2}

z=-1-\sqrt{3}i, z=-2-2\sqrt{3}i

Solutions for Exercise 9 - Polar Form

Determine the Polar Form for each of these Complex numbers

a. z=\sqrt{3}+i

r=\sqrt{(\sqrt{3})^{2}+(1)^{2}}=\sqrt{4}=2

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

z=2(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}})

b. z=3-3i

r=\sqrt{(3)^{2}+(-3)^{2}}=\sqrt{18}=3\sqrt{2}

\theta=\frac{7\pi}{4}

z=3\sqrt{2}(\cos{\frac{7\pi}{4}}+i\sin{\frac{7\pi}{4}})

c. z=1-\sqrt{3}i

r=\sqrt{(1)^{2}+(-\sqrt{3})^{2}}=\sqrt{4}=2

\theta=\frac{5\pi}{3}

z=2(\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}})

d. z=-2i

r=\sqrt{(-2)^{2}}=2

\theta=\frac{3\pi}{2}

z=2(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}})

e. z=-2-2i

r=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{8}=2\sqrt{2}

\theta=\frac{5\pi}{4}

z=2\sqrt{2}(\cos{\frac{5\pi}{4}}+i\sin{\frac{5\pi}{4}})

f. z=-\sqrt{3}+i

r=\sqrt{(-\sqrt{3})^{2}+(1)^{2}}=\sqrt{4}=2

\theta=\frac{5\pi}{6}

z=2(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}})

g. z=4

r=4

\theta=0 or \theta=2\pi

z=4(\cos{0}+i\sin{0}) or z=4(\cos{2\pi}+i\sin{2\pi})

Solutions for Exercise 10 - Roots of Equations

Solve for the roots of these equations

a. (x-3)^{2}=-16

(x-3)^{2}=-16\implies\sqrt{(x-3)^{2}}=\sqrt{-16}

x-3=4i and x=3+4i or x-3=-4i and x=3-4i

solutions are (3+4i,3-4i)

b. (x+iy)^{2}=5-12i

Expand (x+iy)^{2}=x^{2}+2ixy-y^{2}

Set the Real parts and the Imaginary parts of each side equal to each other

x^{2}-y^{2}=5 and 2ixy=-12i\implies xy=-6

If we take the modulus of each side, we can obtain a expression for x^{2}+y^{2}

|(x+iy)^{2}|=|5-12i|=\sqrt{(5)^{2}+(-12)^{2}}=\sqrt{169}=13

Now  |(x+iy)^{2}|=|x+iy|^{2}=(\sqrt{x^{2}+y^{2}})(\sqrt{x^{2}+y^{2}})=x^{2}+y^{2}

and |(x+iy)^{2}|=|5-12i|=13\implies x^{2}+y^{2}=13

Then x^{2}-y^{2}=5\implies y^{2}=x^{2}-5 and x^{2}+y^{2}=x^{2}+x^{2}-5=13\implies 2x^{2}=18 or x^{2}=9 so

x=3 with xy=-6\implies y=-2 or x=-3 with xy=-6\implies y=2

solutions x+iy=3-2i and x+iy=-3+2i

with (3-2i)^{2}=5-12i or (-3+2i)^{2}=5-12i

c. z^{2}=-2i

z^{2}=(x+iy)^{2}=x^{2}-y^{2}+2ixy=-2i

then x^{2}-y^{2}=0\implies x^{2}=y^{2} and 2ixy=-2i\implies xy=-1

with x^{2}+y^{2}=2 with y^{2}=x^{2}\implies x^{2}+x^{2}=2x^{2}=2\implies x^{2}=1

then x=1 with xy=-1\implies y=-1 or x=-1 with xy=-1\implies y=1

solutions x+iy=1-i and x+iy=-1+1

with (1-i)^{2}=-2i or (-1+i)^{2}=-2i

Solutions for Exercise 11 - Powers of a Complex Number

Calculate the following numbers

a. z^{3}=(1-\sqrt{3}i)^{3}

\theta=\frac{5\pi}{3}\implies 3\theta=\frac{15\pi}{3}=5\pi which is equivalent to \theta=\pi

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

z^{3}=2^{3}(\cos{\pi}+i\sin{\pi})=8(-1+0i)=-8

b. \frac{(1+i)^{3}}{(1-i)^{5}}

z_1=(1+i) with \theta_1=\frac{\pi}{4}\implies 3\theta_1=\frac{3\pi}{4} and r_1=\sqrt{(1)^{2}+(1)^{2}}=\sqrt{2}

z_2=(1-i) with \theta_2=\frac{7\pi}{4}\implies 5\theta_2=\frac{35\pi}{4} or 3\theta_2=\frac{3\pi}{4} and r_2=\sqrt{(1)^{2}+(-1)^{2}}=\sqrt{2}

Expand (\cos{\theta}+i\sin{\theta})^{6} to find

Binomial expansion for n=6: 1  6  15  20  15  6  1

(\cos{\theta}+i\sin{\theta})^{6}=x^{6}+6x^{5}iy+15x^{4}i^{2}y^{2}+20x^{3}i^{3}y^{3}+15x^{2}i^{4}y^{4}+6xi^{5}y^{5}+i^{6}y^{6}=x^{6}+6x^{5}iy-15x^{4}y^{2}-20x^{3}iy^{3}+15x^{2}y^{4}+6xiy^{5}-y^{6}

c. \cos{6\theta}

x^{6}-15x^{4}y^{2}+15x^{2}y^{4}-y^{6}

d. \sin{6\theta}

6x^{5}iy-20x^{3}iy^{3}+6xiy^{5}

If \theta=\frac{\pi}{4} find the explicit answer for

e. (\cos{\theta}+i\sin{\theta})^{3}

\theta=\frac{\pi}{4}\implies3\theta=\frac{3\pi}{4}

(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4})^{3}=\cos{\frac{3\pi}{4}}+i\sin{\frac{3\pi}{4}=-1+i

f. (\cos{\theta}+i\sin{\theta})^{4}

\theta=\frac{\pi}{4}\implies4\theta=\frac{4\pi}{4}=\pi

(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4})^{4}=\cos{\pi}+i\sin{\pi}=-1

Solutions for Exercise 12 - Complex Roots

a. Calculate the 4th roots of unity z^{4}=1
z^{n}=\cos{\frac{2k\pi}{n}}+i\sin{\frac{2k\pi}{n}}
with k=0,1...(n-1) where n=4 k=0, 1, 2, 3
z_1=\cos{\frac{2(0)\pi}{4}}+i\sin{\frac{2(0)\pi}{4}}=\cos{0}+i\sin{0}=1+0i=1
z_2=\cos{\frac{2(1)\pi}{4}}+i\sin{\frac{2(1)\pi}{4}}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=0+i=i
z_3=\cos{\frac{2(2)\pi}{4}}+i\sin{\frac{2(2)\pi}{4}}=\cos{\pi}+i\sin{\pi}=-1+0i=-1
z_4=\cos{\frac{2(3)\pi}{4}}+i\sin{\frac{2(3)\pi}{4}}=\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}=0-i=-i
4th Roots of Unity: {1, i, -1, -i}b.
Calculate the 6th roots of unity

    z^{6}=1<span class="ql-right-eqno">   </span><span class="ql-left-eqno">   </span><img src="https://www.superprof.co.uk/resources/wp-content/ql-cache/quicklatex.com-1448b5182dbeeb04cc189e1ea865f46c_l3.png" height="57" width="371" class="ql-img-displayed-equation " alt="\[n=6$ and $k=0, 1, 2, 3, 4, 5\]" title="Rendered by QuickLaTeX.com"/>z_0=\cos{\frac{2(0)\pi}{6}}+i\sin{\frac{2(0)\pi}{6}}=\cos{0}+i\sin{0}=1+oi=1

z_1=\cos{\frac{2(1)\pi}{6}}+i\sin{\frac{2(1)\pi}{6}}=\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}=1+\sqrt{3}i

z_2=\cos{\frac{2(2)\pi}{6}}+i\sin{\frac{2(2)\pi}{6}}=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}=-1+\sqrt{3}i

z_3=\cos{\frac{2(3)\pi}{6}}+i\sin{\frac{2(3)\pi}{6}}=\cos{\pi}+i\sin{\pi}=-1+0i=-1

z_4=\cos{\frac{2(4)\pi}{6}}+i\sin{\frac{2(4)\pi}{6}}=\cos{\frac{4\pi}{3}}+i\sin{\frac{4\pi}{3}}=-1-\sqrt{3}i

z_5=\cos{\frac{2(5)\pi}{6}}+i\sin{\frac{2(5)\pi}{6}}=\cos{\frac{5\pi}{3}}+i\sin{\frac{5\pi}{3}}=1-\sqrt{3}i

6th Roots of Unity: {1, 1+\sqrt{3}i, -1+\sqrt{3}i, -1, -1-\sqrt{3}i, 1-\sqrt{3}i}

c. Calculate (3+4i)^{\frac{1}{2}}

z^{\frac{1}{2}}=\pm[\sqrt{\frac{|z|+x}{2}}+i\sqrt{\frac{|z|-x}{2}}]

|z|=|3+4i|=\sqrt{(3)^{2}+(4)^{2}}=5

z^{\frac{1}{2}}=(3+4i)^{\frac{1}{2}}=\pm[\sqrt{\frac{5+3}{2}}+i\sqrt{\frac{5-3}{2}}]=\pm[\sqrt{4}+i\sqrt{1}]=\pm(2+i)

because

(2+i)^{2}=(2+i)(2+i)=(4-1)+(2+2)i=3+4i and (2+i)^{2}=(-(2+i))^{2}

Calculate the roots of these equations

d. (-1+\sqrt{3})^{\frac{1}{2}}

z^{\frac{1}{2}}=r^{\frac{1}{2}}(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})

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

\theta=\frac{2\pi}{3}\implies \frac{\theta}{2}=\frac{2\pi}{6}=\frac{\pi}{3}

*** QuickLaTeX cannot compile formula: (-1+\sqrt{3})^{\frac{1}{2}}=\sqrt{2}(\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}})=\sqrt{2}(\frac{1}{2}+i\frac{\sqrt{3}}{2})=\frac{\sqrt{2}{2}+\frac{\sqrt{6}}{2}i  *** Error message: File ended while scanning use of \frac .  Emergency stop.   

Use the Complex version of the Quadratic Formula to obtain the roots to the equation z^{2}-(3+2i)z+(1+3i)=0

Quadratic Equation: Az^{2}+Bz+C=0

Quadratic Formula: z=\frac{-B\pm(\sqrt{B^{2}-4AC})}{2A}

z^{2}-(3+2i)z+(1+3i)=0 with A=1, B=-(3+2i)=(-3-2i) and C=(1+3i)

then z=\frac{-(-3-2i)\pm(\sqrt{(-3-2i)^{2}-4(1)(1+3i)})}{2(1)}=z=\frac{(3+2i)\pm(\sqrt{(5+12i)+(-4-12i)})}{2}=

\frac{(3+2i)\pm(\sqrt{1})}{2}=\frac{3+2i\pm1}{2}

z_1=\frac{3+1+2i}{2}=\frac{4+2i}{2}=2+i and z_2=\frac{3-1+2i}{2}=\frac{2+2i}{2}=1+i

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Patrick