Simplify the expression (1+3i)/(1+i)

 Exercise

$\frac{1+3i}{1+i}$

 

Factor the sum or difference of cubes using the formula: $a^3\pm b^3 = (a\pm b)(a^2\mp ab+b^2)$

$\frac{1+3i}{\left(\sqrt[3]{1}+\sqrt[3]{i}\right)\left(\sqrt[3]{\left(1\right)^{2}}-\sqrt[3]{1}\sqrt[3]{i}+\sqrt[3]{i^{2}}\right)}$

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Calculate the power $\sqrt[3]{1}$

$\frac{1+3i}{\left(1+\sqrt[3]{i}\right)\left(\sqrt[3]{\left(1\right)^{2}}-\sqrt[3]{1}\sqrt[3]{i}+\sqrt[3]{i^{2}}\right)}$

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Calculate the power $\sqrt[3]{\left(1\right)^{2}}$

$\frac{1+3i}{\left(1+\sqrt[3]{i}\right)\left(1- 1\sqrt[3]{i}+\sqrt[3]{i^{2}}\right)}$

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Multiply $-1$ times $1$

$\frac{1+3i}{\left(1+\sqrt[3]{i}\right)\left(1-\sqrt[3]{i}+\sqrt[3]{i^{2}}\right)}$

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Cancel like terms $-\sqrt[3]{i}$ and $\sqrt[3]{i^{2}}$

$\frac{1+3i}{\left(1+\sqrt[3]{i}\right)\cdot 1}$

$\frac{1+3i}{\left(1+\sqrt[3]{i}\right)\cdot 1}$

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