Complex number Q.

[1123581321]

not sure if this belongs here,

 

However, i was wondering with: 3+i/1-i = 3+3i+i-1/2 ; in the process of solving it,

Where does the '2' come from..(in the denominator).

[imatfaal]

FibSeq - you gotta either use LaTex or some brackets. Its really difficult to even understand what question you are asking - let alone start to answer it!

 

.

 

Division as follows

[math] \frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i[/math]

 

Assuming yours reads

 

[math] \frac{3+i}{1-i}=\frac{3+3i + i -1}{2}[/math]

 

LHS

[math] \frac{3+i}{1-i}=\frac{3\cdot1+1(-1)}{1^2+(-1)^2}+\frac{1\cdot1-3(-1)}{1^2+(-1)^2}i[/math]

 

[math] \frac{3+i}{1-i}=\frac{2}{2} + \frac{4}{2}i = 1+2i[/math]

 

RHS

 

[math] \frac{3+3i + i -1}{2} = \frac{2+4i}{2}= 1+2i[/math]

[1123581321]

Ok, thanks. But when, one side gets the 'i' and the other doesnt.. is that because the denominators cancels out, since your multiplying it by itself, while the numerator keeps it ?

Also, how exactly are they split to get the LHS & RHS ?

[imatfaal]

Not quite sure what you mean. Division by complex number is defined as I put as the top line of formula

 

[math] \frac{a+bi}{c+di}=\frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i [/math]

 

we can show this as follows

1. multiply both top and bottom by complex conjugate of denominator

 

[math] \frac{a+bi}{c+di}= \frac{(a+bi)(c-di)}{(c+di)(c-di)}[/math]

 

evaluate the top and bottom by multiplying out the two sets of brackets (FOIL)

 

[math] \frac{a+bi}{c+di}= \frac{ac-adi+bci-bidi}{c^2-cdi+cdi-d^2i^2}[/math]

 

Clean up the top and bottom by

1. getting rid of i^2

 

[math] \frac{a+bi}{c+di}= \frac{ac-adi+bci+bd}{c^2-cdi+cdi+d^2}[/math]

 

2. cancelling out terms to give the equation which I guess you used

 

[math] \frac{a+bi}{c+di}= \frac{ac-adi+bci+bd}{c^2+d^2}[/math]

 

3. rearranging real and imaginary

 

[math] \frac{a+bi}{c+di}= \frac{(ac+bd)+ (bc-ad)i}{c^2+d^2}[/math]

 

or as I had it

 

[math] \frac{a+bi}{c+di}= \frac{(ac+bd)}{c^2+d^2} + \frac{(bc-ad)}{c^2+d^2}i[/math]

 

And there you have it! The i cancel outs on the bottom as it is being mutliplied by the complex conjugate

 

Regarding LHS and RHS I just use this because I don't like changing both sides of an equation at the same time - so I took your question and manipulated the Left hand side first to get 1+2i and then rearranged the RHS to show that it is also equal to 1+2i

[1123581321]

thanks, that clears it up.

 

also, how exactly does; | 2-i |^2 = 2^2 + (-1)^2 = 5 ?

 

note; | | , is the abs value.

 

and how does; n as a subset of N, where n^3 > 4 , have an infimum of 3, when clearly 2^3 = 8, which is bigger than 4 ?.

[imatfaal]

What do you understand the absolute value is? I learnt it as a distinct and intuitive part of the number on the complex plane - and your answer makes sense. Take a read of one of the internet math pages on absolute values of complex numbers and it will be clear

[1123581321]

What do you understand the absolute value is? I learnt it as a distinct and intuitive part of the number on the complex plane - and your answer makes sense. Take a read of one of the internet math pages on absolute values of complex numbers and it will be clear

 

Thanks I will...

And would u b able to explain that last Q to me..

 

Thanks I will...

And would u b able to explain that last Q to me..

[1123581321]

Also, how exactly is; (n^2 - 1)/n^3 divergent

 

in my answers, they somehow got; 1/2^n from it. could you please explain..