Let $z = x + y\mathbf{i}$ and $k$ is a real constant. ${{({{z}^{*}})}^{*}}=z$ \(z+{{z}^{*

Question

Let $z = x + y\mathbf{i}$ and $k$ is a real constant. ${{({{z}^{*}})}^{*}}=z$ $z+{{z}^{*}}=2\mathrm{Re}(z)$ $z-z^*=\left(2\mathrm{Im}(z)\right)i$ $z{{z}^{*}}={{x}^{2}}+{{y}^{2}}$ ${{({{z}{1}}\pm {{z}{2}})}^{}}={{z}{1}}^{}\pm {{z}{2}}^{*}$ ${{(kz)}^{}}=k{{z}^{}}$ ${{({{z}_{1}}{{z}_{2}})}^{*}}={{z}_{1}}^{*}{{z}_{2}}^{*}$ ${{\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)}^{*}}=\frac{{{z}_{1}}^{*}}{{{z}_{2}}^{*}}$

Accepted Answer

(a) $(z^*)^*=z$ (b) $z+z^*=2\operatorname{Re}(z)$ (c) $z-z^*=2\operatorname{Im}(z)\mathbf i$ (d) $zz^*=|z|^2=x^2+y^2$ (e) $(z_1\pm z_2)^*=z_1^*\pm z_2^*$ (f) $(kz)^*=kz^*$ (g) $(z_1z_2)^*=z_1^*z_2^*$ (h) $\left(\dfrac{z_1}{z_2}\right)^*=\dfrac{z_1^*}{z_2^*}$, $z_2\ne0$

Property	Identity
Double conjugate	\((z^)^=z\)
Real part	\(z+z^*=2\operatorname{Re}(z)\)
Imaginary part	\(z-z^*=2i\operatorname{Im}(z)\)
Product	\(zz^*=x^2+y^2=\|z\|^2\)
Sum/difference	\((z_1\pm z_2)^=z_1^\pm z_2^*\)
Real scalar	\((kz)^=kz^\), \(k\in\mathbb R\)
Product of numbers	\((z_1z_2)^=z_1^\,z_2^*\)
Quotient	\(\displaystyle\left(\frac{z_1}{z_2}\right)^=\frac{z_1^}{z_2^*}\), \(z_2\ne0\)
Power	\((z^n)^=(z^)^n\)

Properties of Conjugates

Properties of Conjugates

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