But why is it 2z1/2 * z3/2Where did the z3/2 come from? Why are they not combined?
You are allowing yourself to get hung up on something you have known forever because it appears in a new form.
\(\displaystyle (m + n)^2 \equiv (m + n)(m + n) = \)
\(\displaystyle m^2 + mn + nm + n^2 = m^2 + mn + mn + n^2 = m^2 + 2(mn) + n^2.\)
Did we really need to prove that again? Do you really have difficulty grasping where the 2 came from? NO, of course you don't. You are just looking at an old stand by in a new context.
\(\displaystyle \text {Let } m = z^{(1/2)} \text { and } n = z^{(3/2)}.\)
\(\displaystyle \left ( z^{(1/2)} + z^{(3/2)} \right )\left ( z^{(1/2)} + z^{(3/2)} \right ) = (m + n)(m + n) =\)
\(\displaystyle m^2 + 2mn + n^2 = \left ( z^{(1/2)} \right )^2 + 2 \left (z^{(1/2)} * z^{(3/2)} \right ) + \left (z^{(3/2)} \right )^2.\)
Now this of course can be further simplified by combining exponents using the well known laws of exponents.
\(\displaystyle \left ( z^{(1/2)} \right )^2 + 2 \left (z^{(1/2)} * z^{(3/2)} \right ) + \left (z^{(3/2)} \right )^2 = z^{\{2 * (1/2)\}} + 2 * z^{\{(1/2)+ (3/2)\}} + z^{\{2 * (3/2)\}} =\)
\(\displaystyle z^{(2/2)} + 2 * z^{(4/2)} + z^{(6/2)} = z^1 + 2z^2 + z^3 = z + 2z^2 + z^3.\)
Now it might be clearer for you if, as you implied, we simplify exponents first.
\(\displaystyle \left ( z^{(1/2)} + z^{(3/2)} \right )\left ( z^{(1/2)} + z^{(3/2)} \right ) = z^{(1/2)}\left ( z^{(1/2)} + z^{(3/2)} \right ) + z^{(3/2)}\left ( z^{(1/2)} + z^{(3/2)} \right ) =\)
\(\displaystyle \left ( z^{\{(1/2)+(1/2)\}} + z^{\{(1/2)+(3/2)\}} \right ) + \left ( z^{\{(3/2)+(1/2)\}} + z^{\{(3/2)+(3/2)} \right ) =\)
\(\displaystyle (z^{(2/2)} + z^{(4/2)}) + (z^{(4/2)} + z^{6/2)}) = (z^1 + z^2) + (z^2 + z^3) = z + 2z^2 + z^3.\)
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