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The complex numbers solved in the form of systems of equations

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Course
Mathematic

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.

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Uploaded on August 28, 2024
Number of pages 6
Written in 2023/2024
Type Class notes
Professor(s) Gani
Contains All classes

find the complex numbers
complex modulus form
complex conjugate form
the
the complex form for z and w can be written
the complex numbers solved in the form of systems
the real part of the complex form

Institution Secondary school
Mathematic
School year 1

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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

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