100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
The complex numbers solved in the form of systems of equations R360,23   Add to cart

Class notes

The complex numbers solved in the form of systems of equations

 18 views  0 purchase
  • Course
  • Institution

In this document, several tasks of complex numbers have been solved, where for their solution we have used a form of systems of equations. Their solutions are very detailed. Each step and each letter contain its meaning.

Preview 2 out of 6  pages

  • August 28, 2024
  • 6
  • 2023/2024
  • Class notes
  • Gani
  • All classes
  • Secondary school
  • 1
avatar-seller
Math
Complex Numbers
Here we have selected some tasks from complex numbers. The solutions to the
tasks are more detailed. Below we will provide some details to clarify the indexes.



Appendices


Re- The real part
Im- The imaginary part
i- Imaginary number, 𝑖 = √−1
Z- Letter for the complex form
|𝒁| - Complex modulus form
x- The real part of the complex form
i*y – The imaginary part of the complex form
W- Another letter for the complex form
̅̅̅ - Complex conjugate form
𝑾
|𝑾|- Complex modulus form
a- The real part of the complex form for W
i*b- The imaginary part of the complex form for W

, 1. Find the complex numbers Z1 & Z2 for the conditions Re (Z1+Z2) = 3 and
Im (Z1-Z2) = 7, if is Z1= (s - 5*t) + i* (2*s - 3*t) & 𝑍2 = (𝑠 + 2 ∗ 𝑡) +
𝑖 ∗ (𝑠 + 𝑡).
Solution:
Condition: 𝑅𝑒(𝑍1 + 𝑍2 ) = 3 , the real part for 𝑍1 = (s − 5 ∗ t) and
𝑍2 = (s + 2 ∗ t) ;
(𝑠 − 5 ∗ 𝑡) + (s + 2 ∗ t) = 3
𝑠−5∗𝑡+s + 2∗t=3
2𝑠 − 3𝑡 = 3 … (1)
Condition: Im(𝑍1 − Z2 ) = 7, the imaginary part for 𝑍1 = i ∗ (2 ∗ s − 3 ∗
t) and 𝑍2 = i ∗ (s + t);
(2 ∗ s − 3 ∗ t) − (s + t) = 7
2∗s − 3∗t− s− t = 7
𝑠 − 4𝑡 = 7… (2)

2∗𝑠−3∗𝑡 =3 ~ 2∗𝑠−3∗𝑡 =3 ~
𝑠 − 4 ∗ 𝑡 = 7 *(-2) −2 ∗ 𝑠 + 8 ∗ 𝑡 = −14

−11
5 ∗ 𝑡 = −11 𝒕=
5


11
Substitute 𝑡 = − in equation number (2):
5
−11 9
𝑠−4∗( )=7→𝒔=−
5 5


We substitute the result of s and t in Z1 & Z2:

9 −11 −9 −11
Z1 = (− − 5 ∗ ) + i ∗ (2 ∗ − 3∗ ) = 9.2 + 3 ∗ i
5 5 5 5
9 −11 9 11
𝑍2 = (− + 2 ∗ ) + 𝑖 ∗ (− 5 − 5 ) = −6.2 − 4 ∗ i
5 5

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller atdhegashi. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for R360,23. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

67474 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 14 years now

Start selling
R360,23
  • (0)
  Buy now