Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Australia
- University of Technology, Sydney
- Mathematics
- Mathematics 33360
- Various
- Review of Complex Algebra

Jacqui D.

i = ?

√(-1)

A complex number is an expression of what form?

a + bi

Advertisement
)

What is the complex conjugate of x + iy?

x - iy

(a + bi)(x + yi) = ?

(ax - by) + (bx + ay)i

(a + bi)/(c + di) = ?

(ac + bd)/(c^{2} - d^{2})

+ (bc - ad) / (c^{2} - d^{2})i

+ (bc - ad) / (c

3 + 2i

The modulus ιzι of a complex number...

Is the distance of this number from the origin in the Argand diagram

ιzι = √(x^{2} + y^{2})

ιzι = √(x

What is the polar representation of a complex number?

z = r (cosθ + isinθ)

r is the modulus ιzι

θ is the angle between Re axis & r

θ = arctan y/x

r is the modulus ιzι

θ is the angle between Re axis & r

θ = arctan y/x

f(x) = e^{x} has the property such that f(x + y) = ?

f(x)f(y)

Is the complex function periodic?

Yes, it is periodic with a period of 2πi

State Euler's formula

e^{iθ} = cosθ + isinθ

Write z = 2 + 2√3 i in polar form

ιzι = √(4+4x3) = 4

θ = arctan(2√3)/2 = π/3

z = 4( cos(π/3) + isin(π/3))

θ = arctan(2√3)/2 = π/3

z = 4( cos(π/3) + isin(π/3))

Write z=-3i in polar form

ιzι = 3

θ = -π/2

z = 3(cos(π/2) - isin(π/2))

θ = -π/2

z = 3(cos(π/2) - isin(π/2))

Write z = -√6 - √2 i in polar and exponential form

Polar: ιzι = √(6+2) = 2√2

θ = arctan √2/√6 = -5π/6

z = 2√2 (cos(-5π/6) + isin(-5π/6))

Exponential: z = 2√2e^{-i5π/6}

θ = arctan √2/√6 = -5π/6

z = 2√2 (cos(-5π/6) + isin(-5π/6))

Exponential: z = 2√2e

z_{1}= r_{1} (cosθ_{1} + isinθ_{1})

z_{2}= r_{2}(cosθ_{2} + isinθ_{2})

z_{1}z_{2 = ?}

z

z

r_{1}r_{2 }(cos(θ_{1}+θ_{2})) + isin(θ_{1}+θ_{2}))

Advertisement

z_{1}= r_{1} (cosθ_{1} + isinθ_{1})

z_{2}= r_{2}(cosθ_{2} + isinθ_{2})

z_{1 / }z_{2} = ?_{}

z

z

(r_{1}/r_{2} ) (cos(θ_{1}-θ2) + isin(θ_{1}-θ_{2})_{})

What is De Moivre's Theorem?

A theorem useful for developing expressions involving powers of trig. functions and for finding roots of a complex numbers

z^{n} = r^{n}e^{inθ}

z

Fill in the blanks in terms of De Moivre's theorem:

z^{n} = (r(...?))^{n}

= r^{n}(...)

z

= r

z^{n} = (r(cos θ + isin θ))^{n}

z^{n} = r^{n}(cos nθ + isin nθ)

z

What is the complex exponential expression of De Moivre's theorem?

z^{n} = r^{n}e^{inθ}

w

w

Solution shown at www.youtube.com/watch?v=xPn8pHCZL18

w_{0} = 3(cos0 + isin0)

w_{1} = 3(cos2π/3 + isin2π/3)

w_{2} = 3(cos4π/3 + isin4π/3)

w

w

* The material on this site is created by StudyBlue users. StudyBlue is not affiliated with, sponsored by or endorsed by the academic institution or instructor.

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly!??I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU