Mar 7, 2006 #2 benorin Science Advisor Insights Author Messages 1,442 Reaction score 191 Set g(z)=z, that is, write 2 + zi = z and then solve for z.
Mar 7, 2006 #3 meee Messages 87 Reaction score 0 ok.. thanks, any tips in solving for z? i got.. z = 2 + zi z/z = 2/z + i 1 = 2/z + i 2/z = 1 - i z= 2/(1-i) dont think I am going the right way Last edited: Mar 7, 2006
ok.. thanks, any tips in solving for z? i got.. z = 2 + zi z/z = 2/z + i 1 = 2/z + i 2/z = 1 - i z= 2/(1-i) dont think I am going the right way
Mar 7, 2006 #4 benorin Science Advisor Insights Author Messages 1,442 Reaction score 191 2 + iz = z 2 = z -iz = z(1-i) and so, like you have z= \frac{2}{1-i} now multiply by the conjugate of the denominator on top and bottom
2 + iz = z 2 = z -iz = z(1-i) and so, like you have z= \frac{2}{1-i} now multiply by the conjugate of the denominator on top and bottom
Mar 7, 2006 #5 meee Messages 87 Reaction score 0 wel... we can divide 2 by (1 - i) to get z...OHHHhhhhhhhhhhhhhhhhhhhh ahhhhhhhhhh.!1 forgot that!1 multiple by conjugate... how silly of me thankyou so much! !1 yaaay... bottom: (1-i) * (1+i) = 2 top: 2*(1+i) = 2+2i = (2+2i)/2 = 1+i Yayaya Thnnx So mUcH! Last edited: Mar 7, 2006
wel... we can divide 2 by (1 - i) to get z...OHHHhhhhhhhhhhhhhhhhhhhh ahhhhhhhhhh.!1 forgot that!1 multiple by conjugate... how silly of me thankyou so much! !1 yaaay... bottom: (1-i) * (1+i) = 2 top: 2*(1+i) = 2+2i = (2+2i)/2 = 1+i Yayaya Thnnx So mUcH!
Mar 7, 2006 #6 benorin Science Advisor Insights Author Messages 1,442 Reaction score 191 z= \frac{2}{1-i} =\left( \frac{2}{1-i}\right) \left( \frac{1+i}{1+i} \right) = \frac{2(1+i)}{(1-i)(1+i)} = \frac{2(1+i)}{1^2-i^2} = \frac{2(1+i)}{1-(-1)} = 1+i
z= \frac{2}{1-i} =\left( \frac{2}{1-i}\right) \left( \frac{1+i}{1+i} \right) = \frac{2(1+i)}{(1-i)(1+i)} = \frac{2(1+i)}{1^2-i^2} = \frac{2(1+i)}{1-(-1)} = 1+i