B bos1234 Member Joined Oct 9, 2006 Messages 490 Gender Male HSC 2007 Jan 24, 2007 #1 How do i prove that: conjugate(z1+z2) = conjugate(z1) + conjugate(z2)
S SoulSearcher Active Member Joined Oct 13, 2005 Messages 6,751 Location Entangled in the fabric of space-time ... Gender Male HSC 2007 Jan 24, 2007 #2 Let z1 = a + bi and z2 = c + di Then substitute and manipulate until you get LHS = RHS, shouldn't be too hard.
Let z1 = a + bi and z2 = c + di Then substitute and manipulate until you get LHS = RHS, shouldn't be too hard.
B bos1234 Member Joined Oct 9, 2006 Messages 490 Gender Male HSC 2007 Jan 24, 2007 #3 what does conjugate(z1+z2) mean? z1-z2? ------------------------------------------------------ z exists as C such that Re(z) = 2Imz, and z^2 -4i is real. Find z
what does conjugate(z1+z2) mean? z1-z2? ------------------------------------------------------ z exists as C such that Re(z) = 2Imz, and z^2 -4i is real. Find z
P pLuvia Guest Jan 24, 2007 #4 I think that proof is in the Cambridge MX2 textbook, but yes just follow SoulSearcher's method, or you could use: z1=rcis@ z2=tcis# Either way is fine, but I think SS way is quicker
I think that proof is in the Cambridge MX2 textbook, but yes just follow SoulSearcher's method, or you could use: z1=rcis@ z2=tcis# Either way is fine, but I think SS way is quicker
P pLuvia Guest Jan 24, 2007 #5 bos1234 said: what does conjugate(z1+z2) mean? z1-z2? Click to expand... No, if you were to use SS's method a+bi+c+di =(a+c)+i(b+d) Conjugate is =(a+c)-i(b+d)
bos1234 said: what does conjugate(z1+z2) mean? z1-z2? Click to expand... No, if you were to use SS's method a+bi+c+di =(a+c)+i(b+d) Conjugate is =(a+c)-i(b+d)
B bos1234 Member Joined Oct 9, 2006 Messages 490 Gender Male HSC 2007 Jan 24, 2007 #6 ok thanks - its a question in the cmbridge book and i left another qn
