1123581321 Posted September 30, 2012 Share Posted September 30, 2012 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). Link to comment Share on other sites More sharing options...

imatfaal Posted September 30, 2012 Share Posted September 30, 2012 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] Link to comment Share on other sites More sharing options...

1123581321 Posted October 1, 2012 Author Share Posted October 1, 2012 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 ? Link to comment Share on other sites More sharing options...

imatfaal Posted October 1, 2012 Share Posted October 1, 2012 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 Link to comment Share on other sites More sharing options...

1123581321 Posted October 4, 2012 Author Share Posted October 4, 2012 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 ?. Link to comment Share on other sites More sharing options...

imatfaal Posted October 4, 2012 Share Posted October 4, 2012 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 Link to comment Share on other sites More sharing options...

1123581321 Posted October 5, 2012 Author Share Posted October 5, 2012 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.. Link to comment Share on other sites More sharing options...

1123581321 Posted October 6, 2012 Author Share Posted October 6, 2012 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.. Link to comment Share on other sites More sharing options...

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