Comments on: Complex/Imaginary numbers http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/ Maths help and revision for GCSE, A/AS Level and Further Maths Fri, 25 Nov 2011 18:26:31 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: Peter L. Griffiths http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/comment-page-1/#comment-5899 Peter L. Griffiths Wed, 24 Aug 2011 16:58:47 +0000 http://trevorpythag.wordpress.com/?p=57#comment-5899 The n nth roots of the complex number a+ib can be calculated by finding the arcotangent below 90 degrees of the complex ratio a/b and dividing it by n the root required. The cotangent of the result will be one of the roots. The remaining n-1 roots can be calculated by adding 360 degrees on to the arcotangent until the n nth roots are calculated. A similar procedure is involved for computing the n nth roots of +1, -1, i and -i. a is made to equal 0 so that the Cotes format can be applied being cos90+isin90 equals 0+i, also cos180+isin180 equals -1 and cos360+isin360 equals +1 which is i to the power of 4. This system also applies for division, cos45+isin45 equals the square root of i. The n nth roots of the complex number a+ib can be calculated by finding the arcotangent below 90 degrees of the complex ratio a/b and dividing it by n the root required. The cotangent of the result will be one of the roots. The remaining n-1 roots can be calculated by adding 360 degrees on to the arcotangent until the n nth roots are calculated. A similar procedure is involved for computing the n nth roots of +1, -1, i and -i. a is made to equal 0 so that the Cotes format can be applied being cos90+isin90 equals 0+i, also cos180+isin180 equals -1 and cos360+isin360 equals +1 which is i to the power of 4. This system also applies for division, cos45+isin45 equals the square root of i.

]]> By: admin http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/comment-page-1/#comment-263 admin Wed, 21 Apr 2010 20:10:38 +0000 http://trevorpythag.wordpress.com/?p=57#comment-263 <a href="#comment-256" rel="nofollow">@Jack </a> Your basically looking for a equation containing only u and v so we need to eliminate the x and y. Firstly take the equation you have with u and v in it and expand it out to get $latex u+iv=x^2 +2xyi - y^2 $ remember that the i makes the y terms squared negative. Now we substitute y=x-1 into this equation, since that is what the other complex number satisfies, so we have an equation in terms of u,v and x.This gives: $latex u + iv = (2x^2 -2)i + 2x -1 We now can compare the real and imaginary terms of each side of the equation to get $latex u = 2x - 1 \Rightarrow x = \frac{u+1}{2} $ and $latex v = 2x^2 -2 $ Substituting the equation for x in terms of u into the equation for v get $latex v = \frac{2(u^2 + 2u + 1)}{4} - 2$ which you can simplify to get $latex v = \frac{1}{2}u^2 + u - \frac{3}{2} $ You may need to check my algebra but the basic method is: >get a simple equation in terms of u,v and x or y or both >use the locus of the other complex number to find an expression for either x or y >substitute this into you equation containing u and v so that this only has one other variable, say x >compare real and imaginary parts to gain two equations of real numbers in terms of u,v and x >use these two to eliminate x (make x the subject of one and substitute into the other) >simplify the resulting expression @Jack
Your basically looking for a equation containing only u and v so we need to eliminate the x and y. Firstly take the equation you have with u and v in it and expand it out to get
$latex u+iv=x^2 +2xyi – y^2 $ remember that the i makes the y terms squared negative. Now we substitute y=x-1 into this equation, since that is what the other complex number satisfies, so we have an equation in terms of u,v and x.This gives:
$latex u + iv = (2x^2 -2)i + 2x -1

We now can compare the real and imaginary terms of each side of the equation to get
$latex u = 2x – 1 \Rightarrow x = \frac{u+1}{2} $ and
$latex v = 2x^2 -2 $
Substituting the equation for x in terms of u into the equation for v get
$latex v = \frac{2(u^2 + 2u + 1)}{4} – 2$
which you can simplify to get
$latex v = \frac{1}{2}u^2 + u – \frac{3}{2} $

You may need to check my algebra but the basic method is:
>get a simple equation in terms of u,v and x or y or both
>use the locus of the other complex number to find an expression for either x or y
>substitute this into you equation containing u and v so that this only has one other variable, say x
>compare real and imaginary parts to gain two equations of real numbers in terms of u,v and x
>use these two to eliminate x (make x the subject of one and substitute into the other)
>simplify the resulting expression

]]> By: Jack http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/comment-page-1/#comment-256 Jack Wed, 14 Apr 2010 17:10:00 +0000 http://trevorpythag.wordpress.com/?p=57#comment-256 What do you do if you are given the relationship between two complex numbers and the equation of the locus of one, and you are asked to find the locus of the other? For example u+iv=(x+iy)^2 y=x-1 Find the locus of u+iv What do you do if you are given the relationship between two complex numbers and the equation of the locus of one, and you are asked to find the locus of the other? For example
u+iv=(x+iy)^2
y=x-1
Find the locus of u+iv

]]> By: trevorpythag http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/comment-page-1/#comment-102 trevorpythag Sat, 06 Dec 2008 17:37:04 +0000 http://trevorpythag.wordpress.com/?p=57#comment-102 you can treat them like you would in normal algebra but whenever you get a i squared you can chnage it to -1 ie in your example the answer would be <strong>(-4i)(-5i)(3i) = -60i</strong> beacuse 4x5x3 = 20 and i x i x i = -i hope that helps dave you can treat them like you would in normal algebra but whenever you get a i squared you can chnage it to -1
ie in your example the answer would be
(-4i)(-5i)(3i) = -60i
beacuse 4x5x3 = 20 and i x i x i = -i
hope that helps dave

]]> By: aroosa http://trevorpythag.co.uk/2008/mathematics/algebra/compleximaginary-numbers/comment-page-1/#comment-101 aroosa Sat, 06 Dec 2008 00:57:26 +0000 http://trevorpythag.wordpress.com/?p=57#comment-101 how do i simply multiply imaginary numbers? like; (-4i)(-5i)(3i) how do i simply multiply imaginary numbers?

like;

(-4i)(-5i)(3i)

]]>