Question

Friends, is this right?
Simplify the quantities in Exercise using
m(z) = z^2.
1) m(z+h)-m(z-h)
(z^2+h)-(z^2-h)
=0

Answer 1

m(z+h)-m(z-h)
\[(z^2+h)-(z^2-h) = 0\]

Answer 2

\(m(z) = z^2\)

\(m(z + h) - m(z - h)\)

\(= (z + h)^2 - (z - h)^2\)

\(= z^2 + 2hz + h^2 - z^2 + 2hz - h^2\)

\(= 4hz\)

Answer 3

@mathstudent55 How did we get the square root outside the brackets? I appreciate your help.

Answer 4

In this problem, there are no square roots.
there is an exponent of 2.

Answer 5

oh sorry, exponent :)))

Answer 6

This is how I interpreted this problem.
The line m(z) = z^2 means function m of variable z is defined as z^2.

Answer 7

Then you are asked to do m(z + h) - m(z - h)
This means evaluate function m at z + h, and subtract from it function m evaluated at z - h.

Answer 8

Since function m(z) equals z^2, then function m(z) evaluated at z + h means to replace z with z + h.
You get:
m(z) = z^2
m(z + h) = (z + h)^2

Similarly, you must evaluate function m(z) at z - h, so replace z with z - h.
You get:
m(z) = z^2
m(z - h) = (z - h)^2

This means that
m(z + h) - m(z - h) = (z + h)^2 - (z - h)^2

Now you need to square the binomials z + h and z - h, and do the subtraction.

(z + h)^2 - (z - h)^2 =

= z^2 + 2hz + h^2 - (z^2 - 2hz + h^2)

= z^2 + 2hz + h^2 - z^2 + 2hz - h^2

= 4hz

Answer 9

Sometimes color helps if you're finding this too confusing c:\[\large\rm m(\color{orangered}{z})=(\color{orangered}{z})^2\]So then,\[\large\rm m(\color{orangered}{z+h})=(\color{orangered}{z+h})^2\]

Answer 10

Just trying to further explain why the square is on the outside c:

Answer 11

do you have an m(h) ?

Answer 12

m(z+h) = (z+h)² = z² + 2zh + h²
m(z-h) = (z-h)² = z² - 2zh + h²

m(z+h) - m(z-h) = z² + 2zh + h² - (z² - 2zh + h²)
= 4zh