Engineering How does complex multiplication work?

[lee123456789]

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

 

[Merlin3189]

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

 

[Mark44]

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.

v = (2xj6)(3-j8)

The right side is clearer as 2(j6)(3 - j8}

2x3=6
2x-j8=-16

2(-j8)

j6x3=j18
j6x-j8=-j48

As you have written the above, it looks like you're subtracting j8 from j6x.
Better as j6(-j8)

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

 

[berkeman]

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!

 

[Merlin3189]

And my eys must be going. I thought it was + !

 

[PeroK]

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.

 

[Tom.G]

How does complex multiplication work?.
Simple answer:
They are just binomials.
So multiply them the same way you would multiply algebraic binomials.

Cheers,
Tom