How does complex multiplication work?

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Discussion Overview

The discussion centers around the process of complex multiplication, specifically in the context of determining a voltage expressed as a complex number. Participants are examining the calculations and clarifying the arithmetic involved, while also addressing the notation used in the problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant presents a calculation for the voltage v = (2xj6)(3-j8) but questions if their approach is correct.
  • Another participant agrees with some parts of the calculation but disputes specific steps, indicating that the notation used is confusing.
  • A third participant suggests that using parentheses instead of 'x' for multiplication would improve clarity in the calculations.
  • There is a mention of potential typos in the original post, leading to confusion about the arithmetic.
  • One participant summarizes that complex numbers can be treated as binomials and suggests multiplying them similarly to algebraic binomials.

Areas of Agreement / Disagreement

Participants express disagreement regarding the clarity and correctness of the original calculations. There is no consensus on the specific errors, as some participants identify issues while others suggest they may be typos. The discussion remains unresolved regarding the exact nature of the mistakes in the calculations.

Contextual Notes

Participants note that the notation used may lead to misunderstandings, particularly the use of 'x' for multiplication. There are also unresolved aspects regarding the arithmetic steps taken in the calculations.

lee123456789
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Homework Statement
Complex Arithmetics Clarification
Relevant Equations
see below
is this right
Q) Determine this voltage in its simplest complex number form.

v = (2xj6)(3-j8)

2x3=6
2x-j8=-16

j6x3=j18
j6x-j8=-j48

v=6 +(j18-j16) - J(^2)48 (j^2 = -1)
v=6 +j2 +48
V=54 + j2
 
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v = (2xj6)(3-j8)

2x3=6 yes
2x-j8=-16 no

j6x3=j18 yes
j6x-j8=-j48 no

v=6 +(j18-j16) - J(^2)48 but this is right! (j^2 = -1)
v=6 +j2 +48
V=54 + j2 yes
 
lee123456789 said:
Homework Statement:: Complex Arithmetics Clarification
Relevant Equations:: see below

is this right
Q) Determine this voltage in its simplest complex number form.
Your work is very difficult to read. At levels of mathematics above arithmetic, 'x' is used as the name of an unknown variable, not multiplication. Instead of using 'x', it's clearer to use parentheses.
lee123456789 said:
v = (2xj6)(3-j8)
The right side is clearer as 2(j6)(3 - j8}
lee123456789 said:
2x3=6
2x-j8=-16
2(-j8)
lee123456789 said:
j6x3=j18
j6x-j8=-j48
As you have written the above, it looks like you're subtracting j8 from j6x.
Better as j6(-j8)
lee123456789 said:
v=6 +(j18-j16) - J(^2)48 (j^2 = -1)
v=6 +j2 +48
V=54 + j2
 
Mark44 said:
Your work is very difficult to read. At levels of mathematics above arithmetic, 'x' is used as the name of an unknown variable, not multiplication.
Oh lordy, no wonder I couldn't figure out what he was doing. I assumed x was a variable as you say. Whew!
 
And my eys must be going. I thought it was + !
 
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Merlin3189 said:
And my eys must be going. I thought it was + !
Well, ##(2+j6)(3-j8) = 54 + j2##

Perhaps the errors in the OP were all typos.
 
How does complex multiplication work?.
Simple answer:
They are just binomials.
So multiply them the same way you would multiply algebraic binomials.

Cheers,
Tom
 

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