How do you prove that when |z|=1, the conjugate of z is equal to 1/z.
(complex numbers)
z = a+bi
|z|=1 ==> a^2+b^2 = 1
1/(a+bi) = (a-bi)/(a^2+b^2) = a-bi = z*
To prove the statement that when |z|=1, the conjugate of z is equal to 1/z, we can use the properties of complex numbers and some algebraic manipulation. Here's how you can approach it:
1. Start with a complex number z = a + bi, where 'a' and 'b' are real numbers representing the real and imaginary parts of z, respectively.
2. Assume that |z| = 1. This means that the magnitude of z, denoted by |z|, is equal to 1. In other words, |z| = √(a^2 + b^2) = 1.
3. Use the definition of the conjugate of a complex number. The conjugate of z, denoted by z*, is obtained by changing the sign of the imaginary part of z. So, z* = a - bi.
4. Express 1/z in terms of z*. This can be done by multiplying the numerator and denominator of 1/z by z*.
(1/z) = (1/z) * (z*/z*) = (z*/|z|^2) = (z*/(a^2 + b^2))
5. Now, compare the expression for z* from step 3 with the expression for 1/z from step 4. Notice that they are equal: z* = (z*/(a^2 + b^2)).
6. Since z* = (z*/(a^2 + b^2)) when |z|=1, we have successfully proved the statement.
Therefore, when |z|=1, the conjugate of z is equal to 1/z.
Note: The proof assumes that z is a nonzero complex number (z ≠ 0), as division by zero is undefined.