prove: if z is a complex number and 1/z=the conjugate of z over the absolute value of z squared.
You can write z in the form:
z = |z|exp(i theta)
is that how you prove it?
Well, if z = |z| exp(i theta), then
1/z = |z|^-1 exp(- i theta)
But z* = z exp(- i theta), so it differs by a factor 1/|z|^2
To prove the given equality, we will manipulate both sides of the equation separately. Let's start by simplifying the right-hand side (RHS) of the equation.
RHS: (conjugate of z) / |z|^2
Step 1: Express the conjugate of z
The conjugate of a complex number z, denoted as z*, is obtained by changing the sign of its imaginary part (i.e., replacing i with -i). So, the conjugate of z is written as z* = a - bi, where z = a + bi.
RHS: (a - bi) / |z|^2
Step 2: Express the absolute value of z squared
The absolute value of a complex number z, denoted as |z|, is given by |z| = √(a^2 + b^2). Squaring this expression will yield the absolute value of z squared, |z|^2 = a^2 + b^2.
RHS: (a - bi) / (a^2 + b^2)
Now, let's focus on the left-hand side (LHS) of the equation and simplify it to match the RHS expression.
LHS: 1 / z
Step 3: Express z in terms of a and b
The complex number z can be written as z = a + bi.
LHS: 1 / (a + bi)
Step 4: Rationalizing the denominator
To simplify the expression further, we need to rationalize the denominator. This is achieved by multiplying both the numerator and denominator by the conjugate of the denominator.
LHS: (1 / (a + bi)) * (a - bi) / (a - bi)
Using the product of conjugates rule (a + bi)(a - bi) = a^2 + b^2, we can simplify the denominator:
LHS: (a - bi) / (a^2 + b^2)
As both the RHS and LHS expressions have been simplified to the same form, we have successfully proven the equality:
(a - bi) / (a^2 + b^2) = (a - bi) / |z|^2
Therefore, if z is a complex number, the equation holds true for any values of a and b, which confirms the validity of the given statement.