Complex Numbers (1 Viewer)

[zeek]

Prove: z²₁ + z²₂=z₁z₂

Show that ||z₁|-|z₂||<=|z₁+z₂|

These are associated with vectors so i think you have to use them

 

[sasquatch]

For the first question you didnt write it properly, lucky i remember it being from the cambridge book:

Note: See the image attached
View attachment 13802

Show z₁² z₂² = z₁ z₂

Let arg(z₁) = a, arg(z₂) = b

From diagram b = a + (π/3)

LHS = |z₁|² cis2a + |z₂|² cis2b
= |z₁|² cis(2a) + |z₁|² cis2b
= |z₁|²( cis(2a) + cis(2b))
= |z₁|²( cis(2a) + cis(2π/3 + 2a))
= |z₁|²( cos(2a) + isin(2a) + cos(2π/3 + 2a) + isin(2π/3 + 2a))
= |z₁|²( cos(2a) + isin(2a) + cos(2π/3)cos(2a) - sin(2π/3)sin(2a) + i[sin(2π/3)cos(2a) + cos(2π/3)sin(2a)])
= |z₁|²( cos(2a) + isin(2a) + (-1/2)cos(2a) - (√3/2)sin(2a) + i[(√3/2)cos(2a) + (-1/2)sin(2a)])
= |z₁|²((1/2)cos(2a) + (1/2)isin(2a) + - (√3/2)sin(2a) + (√3/2)icos(2a))
= |z₁|²((1/2)cos(2a) - (√3/2)sin(2a) + i[(1/2)sin(2a) + (√3/2)cos(2a)])
= |z₁|²(cos(π/3)cos(2a) - sin(π/3)sin(2a) + i[cos(π/3)sin(2a) + sin(π/3)cos(2a)])
= |z₁|²(cos(π/3+2a) + isin(π/3+2a))
= |z₁||z₂|(cos(π/3+a+a) + isin(π/3+a+a))
= |z₁||z₂|(cos(a+b) + isin(a+b))
= |z₁||z₂|cis(a+b)
= z₁z₂
=RHS

therefore z₁² z₂² = z₁ z₂

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I dunno if i did question two correctly, but this is what i did anyway..
View attachment 13803

 

[zeek]

no... the question is right go to page 54 question 5 in cambridge...

thx though

 

[NickP101]

Strangely enough im pretty sure he did answer your question lol even though he said he was answering a different one

 

[zeek]

oh yeah he did lol i didn't look at the working out

 

[Mill]

Zeek, the first question you asked only makes sense in the context of the diagram.

It is not something that is always true.

It is only true for the vectors shown in the diagram (and a few other cases).

 

[zeek]

kewl thx, what about the second one?

 

[Gecko888]

Question 2: I believe this is just basic triangular equality. Let B represent |z1|, C represent |z1+z2| and A represent the origin. Now, Length BC is given by |z2|. So, applying the triangular inequality, |z1+z2|+|z2|>|z1|, so |z1+z2|>|z1|-|z2| (sum of two sides of a triangle is greater than other one.) Also note that |z1+z2|>|z2|-|z1|, so |z1+z2|>||z1|-|z2||. Note although we tend to associate || with modulus, |z1| and |z2| are just real numbers so the second pair of || is just an absolute value sign.

 

[sasquatch]

no... the question is right go to page 54 question 5 in cambridge...

thx though

Strangely enough im pretty sure he did answer your question lol even though he said he was answering a different one

You people didnt read my post properly. I said you didnt write it properly meaning something similar to what Mill said. You cant just prove Q1, by itself you need to know the information saying that the triangle is an Equilateral Triangle. I said that i remember where the question was from, so i could look at it to see the complete quesiton.....