# Exercise ID146

Algebra → Komplex százmok → Komplex számok összeadása-kivonása
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Calculate the following expression if ${z}_{1}=\sqrt{2}-\sqrt{3}i$ and ${z}_{2}=\sqrt{2}+\sqrt{3}i$

${z}_{1}+{z}_{2}\phantom{\rule{0ex}{0ex}}$ ${z}_{1}+{z}_{2}=\sqrt{2}{-}\sqrt{3}{i}+\sqrt{2}{+}\sqrt{3}{i}$ $=2\sqrt{2}$ ${z}_{1}-{z}_{2}\phantom{\rule{0ex}{0ex}}$ ${z}_{1}-{z}_{2}=\sqrt{2}-\sqrt{3}i-\left(\sqrt{2}+\sqrt{3}i\right)$ $=\sqrt{2}{-}\sqrt{2}-\sqrt{3}i-\sqrt{3}i$ $=-2\sqrt{3}i$ ${z}_{1}·{z}_{2}\phantom{\rule{0ex}{0ex}}$ ${z}_{1}·{z}_{2}=\left(\sqrt{2}-\sqrt{3}i\right)\left(\sqrt{2}+\sqrt{3}i\right)$ $=\left(\sqrt{2}{\right)}^{2}-\left(\sqrt{3}i{\right)}^{2}$ $=5$ $\frac{{z}_{1}}{{z}_{2}}\phantom{\rule{0ex}{0ex}}$ $\frac{{z}_{1}}{{z}_{2}}=\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}+\sqrt{3}i}$ $=\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}+\sqrt{3}i}·\frac{\sqrt{2}-\sqrt{3}i}{\sqrt{2}-\sqrt{3}i}$ $=\frac{{\left(\sqrt{2}-\sqrt{3}i\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{3}i\right)}^{2}}$ $=\frac{2-2\sqrt{2}·\sqrt{3}i+{\left(\sqrt{3}i\right)}^{2}}{2-\left(-3\right)}$ $=\frac{2-2\sqrt{6}i-3}{5}$ $=\frac{-1-2\sqrt{6}i}{5}$ $=\frac{-1}{5}-\frac{2\sqrt{6}i}{5}$
${z}_{1}+{z}_{2}=2\sqrt{2}{z}_{1}-{z}_{2}=-2\sqrt{3}i{z}_{1}·{z}_{2}=5\frac{{z}_{1}}{{z}_{2}}=\frac{-1}{5}-\frac{2\sqrt{6}i}{5}$

$i=\sqrt{-1};{i}^{2}=-1$

${a}^{2}-{b}^{2}=\left(a-b\right)\left(a+b\right)$