Calculate the value of the expression: 1^(°) (z-bar…
Calculate the value of the expression: $$ 1^{\circ}\,\, \frac{z-\bar z}{1+z\bar z}, \text{ if } z=1+i; \qquad 2^{\circ}\,\, \frac{z+\bar z}{2z+3}, \text{ if } z=\frac{i-1}{2}. $$
Explanation of Method
For $z=x+iy$, the identities $z-\bar z=2iy$ (purely imaginary), $z+\bar z=2x$ (purely real), and $z\bar z=|z|^2=x^2+y^2$ (real, non-negative) let each numerator be computed instantly. What remains is a straightforward division by a complex denominator, done by multiplying by its conjugate.
Part $1^{\circ}$ · $\dfrac{z-\bar z}{1+z\bar z}$ for $z=1+i$
Step 1 · Find $\bar z$
$$ z=1+i \quad\Longrightarrow\quad \bar z = 1-i $$
Step 2 · Compute $z-\bar z$ and $z\bar z$
$$ z-\bar z = (1+i)-(1-i) = 2i $$
$$ z\bar z = (1+i)(1-i) = 1-i^2 = 2 $$
Step 3 · Assemble the fraction
$$ 1+z\bar z = 1+2 = 3 \quad\Longrightarrow\quad \frac{z-\bar z}{1+z\bar z} = \frac{2i}{3} $$
Answer for $1^{\circ}$
$$ \boxed{\frac{z-\bar z}{1+z\bar z} = \frac{2}{3}i} \qquad (\text{Re}=0,\ \text{Im}=\tfrac23) $$
In general $z-\bar z=2iy$ is always purely imaginary, so whenever the denominator $1+z\bar z$ is real (as it always is), the whole expression is purely imaginary — consistent with the result here.
Part $2^{\circ}$ · $\dfrac{z+\bar z}{2z+3}$ for $z=\dfrac{i-1}{2}$
Step 1 · Write $z$ and $\bar z$ in standard form
$$ z = \frac{-1+i}{2} = -\frac12+\frac12i \quad\Longrightarrow\quad \bar z = -\frac12-\frac12i $$
Step 2 · Compute $z+\bar z$
$$ z+\bar z = \left(-\frac12+\frac12i\right)+\left(-\frac12-\frac12i\right) = -1 $$
(as expected, since $z+\bar z=2\,\text{Re}(z)=2\left(-\frac12\right)=-1$).
Step 3 · Compute $2z+3$
$$ 2z = -1+i \quad\Longrightarrow\quad 2z+3 = 2+i $$
Step 4 · Form and rationalize the fraction
$$ \frac{z+\bar z}{2z+3} = \frac{-1}{2+i} \cdot \frac{2-i}{2-i} = \frac{-(2-i)}{(2+i)(2-i)} = \frac{-2+i}{5} $$
Step 5 · Simplify
$$ \frac{-2+i}{5} = -\frac25+\frac15i $$
Answer for $2^{\circ}$
$$ \boxed{\frac{z+\bar z}{2z+3} = -\frac25+\frac15i} \qquad (\text{Re}=-\tfrac25,\ \text{Im}=\tfrac15) $$
| Identity | Form for $z=x+iy$ |
|---|---|
| $z-\bar z$ | $2iy$ — purely imaginary |
| $z+\bar z$ | $2x$ — purely real |
| $z\bar z$ | $x^2+y^2=|z|^2$ — real, non-negative |
Result
Both results follow immediately once $z\pm\bar z$ and $z\bar z$ are expressed in terms of $\text{Re}(z)$, $\text{Im}(z)$, and $|z|$.