Answer :
Let's break the problem into parts and solve it step-by-step.
### Part 1: Show that [tex]\(\frac{1}{z} = \frac{\bar{z}}{|z|^2}\)[/tex]
Let [tex]\( z = a + bi \)[/tex] be a complex number where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers and [tex]\(i\)[/tex] is the imaginary unit.
1. Complex Conjugate:
The complex conjugate of [tex]\( z \)[/tex], denoted by [tex]\(\bar{z}\)[/tex], is:
[tex]\[ \bar{z} = a - bi \][/tex]
2. Magnitude of z:
The magnitude (or modulus) of [tex]\(z\)[/tex], denoted by [tex]\(|z|\)[/tex], is:
[tex]\[ |z| = \sqrt{a^2 + b^2} \][/tex]
Thus, the square of the magnitude is:
[tex]\[ |z|^2 = a^2 + b^2 \][/tex]
3. Reciprocal of z:
We need to find the reciprocal of [tex]\(z\)[/tex], which is [tex]\(\frac{1}{z}\)[/tex].
Recall the property of complex numbers that [tex]\(\frac{1}{z}\)[/tex] can be expressed using its conjugate and magnitude:
[tex]\[ \frac{1}{z} = \frac{1}{a+bi} \][/tex]
To simplify, multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = \frac{a - bi}{(a+bi)(a-bi)} = \frac{a - bi}{a^2 + b^2} \][/tex]
Therefore,
[tex]\[ \frac{1}{z} = \frac{a - bi}{a^2 + b^2} = \frac{\bar{z}}{|z|^2} \][/tex]
Thus, we have shown that:
[tex]\[ \frac{1}{z} = \frac{\bar{z}}{|z|^2} \][/tex]
### Part 2: Show that [tex]\(\bar{z} = -z\)[/tex] implies [tex]\(z\)[/tex] has only an imaginary part
Given [tex]\(\bar{z} = -z\)[/tex]:
[tex]\[ a - bi = - (a + bi) \][/tex]
Equating real and imaginary parts, we get:
[tex]\[ a = -a \quad \text{and} \quad -bi = -bi \][/tex]
From [tex]\(a = -a\)[/tex], we get [tex]\( a = 0 \)[/tex]. Thus, [tex]\(z\)[/tex] becomes purely imaginary, [tex]\( z = bi \)[/tex].
### Part 3: Solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Given the equation:
[tex]\[ z(2 - i) = (\bar{z} + 1)(1 + i) \][/tex]
We need to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that this equation is satisfied.
First, express [tex]\(z\)[/tex] and [tex]\(\bar{z}\)[/tex]:
[tex]\[ z = a + bi \quad \text{and} \quad \bar{z} = a - bi \][/tex]
Substitute these into the given equation:
[tex]\[ (a + bi)(2 - i) = ((a - bi) + 1)(1 + i) \][/tex]
Expand both sides:
1. Left-hand Side:
[tex]\[ (a + bi)(2 - i) = 2a + 2bi - ai - b i^2 = 2a + 2bi - ai + b = (2a + b) + (2b - a)i \][/tex]
2. Right-hand Side:
[tex]\[ ((a - bi) + 1)(1 + i) = (a - bi + 1)(1 + i) = (a + 1 - bi)(1 + i) \][/tex]
Expand:
[tex]\[ (a + 1 - bi)(1 + i) = (a + 1)(1 + i) - bi(1 + i) \][/tex]
[tex]\[ = (a + 1) + (a + 1)i - bi - b i^2 \][/tex]
[tex]\[ = a + 1 + (a + 1)i - bi + b \][/tex]
[tex]\[ = (a + 1 + b) + (a - b + 1)i \][/tex]
Now equate the real and imaginary parts from both sides:
[tex]\[ 2a + b = a + 1 + b \quad \text{(real part)} \][/tex]
[tex]\[ 2b - a = a - b + 1 \quad \text{(imaginary part)} \][/tex]
From the real part equation:
[tex]\[ 2a + b = a + 1 + b \][/tex]
[tex]\[ 2a = a + 1 \][/tex]
[tex]\[ a = 1 \][/tex]
From the imaginary part equation:
[tex]\[ 2b - a = a - b + 1 \][/tex]
[tex]\[ 2b - 1 = 1 - b + 1 \][/tex]
[tex]\[ 2b - 1 = 2 - b \][/tex]
[tex]\[ 3b = 3 \][/tex]
[tex]\[ b = 1 \][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy the given equation are:
[tex]\[ a = 1, \quad b = 1 \][/tex]
Hence [tex]\( z = 1 + i \)[/tex].
### Part 1: Show that [tex]\(\frac{1}{z} = \frac{\bar{z}}{|z|^2}\)[/tex]
Let [tex]\( z = a + bi \)[/tex] be a complex number where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are real numbers and [tex]\(i\)[/tex] is the imaginary unit.
1. Complex Conjugate:
The complex conjugate of [tex]\( z \)[/tex], denoted by [tex]\(\bar{z}\)[/tex], is:
[tex]\[ \bar{z} = a - bi \][/tex]
2. Magnitude of z:
The magnitude (or modulus) of [tex]\(z\)[/tex], denoted by [tex]\(|z|\)[/tex], is:
[tex]\[ |z| = \sqrt{a^2 + b^2} \][/tex]
Thus, the square of the magnitude is:
[tex]\[ |z|^2 = a^2 + b^2 \][/tex]
3. Reciprocal of z:
We need to find the reciprocal of [tex]\(z\)[/tex], which is [tex]\(\frac{1}{z}\)[/tex].
Recall the property of complex numbers that [tex]\(\frac{1}{z}\)[/tex] can be expressed using its conjugate and magnitude:
[tex]\[ \frac{1}{z} = \frac{1}{a+bi} \][/tex]
To simplify, multiply the numerator and the denominator by the conjugate of the denominator:
[tex]\[ \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} = \frac{a - bi}{(a+bi)(a-bi)} = \frac{a - bi}{a^2 + b^2} \][/tex]
Therefore,
[tex]\[ \frac{1}{z} = \frac{a - bi}{a^2 + b^2} = \frac{\bar{z}}{|z|^2} \][/tex]
Thus, we have shown that:
[tex]\[ \frac{1}{z} = \frac{\bar{z}}{|z|^2} \][/tex]
### Part 2: Show that [tex]\(\bar{z} = -z\)[/tex] implies [tex]\(z\)[/tex] has only an imaginary part
Given [tex]\(\bar{z} = -z\)[/tex]:
[tex]\[ a - bi = - (a + bi) \][/tex]
Equating real and imaginary parts, we get:
[tex]\[ a = -a \quad \text{and} \quad -bi = -bi \][/tex]
From [tex]\(a = -a\)[/tex], we get [tex]\( a = 0 \)[/tex]. Thus, [tex]\(z\)[/tex] becomes purely imaginary, [tex]\( z = bi \)[/tex].
### Part 3: Solve for [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
Given the equation:
[tex]\[ z(2 - i) = (\bar{z} + 1)(1 + i) \][/tex]
We need to find [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that this equation is satisfied.
First, express [tex]\(z\)[/tex] and [tex]\(\bar{z}\)[/tex]:
[tex]\[ z = a + bi \quad \text{and} \quad \bar{z} = a - bi \][/tex]
Substitute these into the given equation:
[tex]\[ (a + bi)(2 - i) = ((a - bi) + 1)(1 + i) \][/tex]
Expand both sides:
1. Left-hand Side:
[tex]\[ (a + bi)(2 - i) = 2a + 2bi - ai - b i^2 = 2a + 2bi - ai + b = (2a + b) + (2b - a)i \][/tex]
2. Right-hand Side:
[tex]\[ ((a - bi) + 1)(1 + i) = (a - bi + 1)(1 + i) = (a + 1 - bi)(1 + i) \][/tex]
Expand:
[tex]\[ (a + 1 - bi)(1 + i) = (a + 1)(1 + i) - bi(1 + i) \][/tex]
[tex]\[ = (a + 1) + (a + 1)i - bi - b i^2 \][/tex]
[tex]\[ = a + 1 + (a + 1)i - bi + b \][/tex]
[tex]\[ = (a + 1 + b) + (a - b + 1)i \][/tex]
Now equate the real and imaginary parts from both sides:
[tex]\[ 2a + b = a + 1 + b \quad \text{(real part)} \][/tex]
[tex]\[ 2b - a = a - b + 1 \quad \text{(imaginary part)} \][/tex]
From the real part equation:
[tex]\[ 2a + b = a + 1 + b \][/tex]
[tex]\[ 2a = a + 1 \][/tex]
[tex]\[ a = 1 \][/tex]
From the imaginary part equation:
[tex]\[ 2b - a = a - b + 1 \][/tex]
[tex]\[ 2b - 1 = 1 - b + 1 \][/tex]
[tex]\[ 2b - 1 = 2 - b \][/tex]
[tex]\[ 3b = 3 \][/tex]
[tex]\[ b = 1 \][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] that satisfy the given equation are:
[tex]\[ a = 1, \quad b = 1 \][/tex]
Hence [tex]\( z = 1 + i \)[/tex].
