# What is 12(cos 60 + i sin 60)

Question
Complex numbers
What is $$\displaystyle{12}{\left({\cos{{60}}}+{i}{\sin{{60}}}\right)}$$

2020-10-22
$$\displaystyle{12}{\left({\cos{{60}}}+{i}{\sin{{60}}}\right)}={12}{\left(\frac{{1}}{{2}}+{i}\frac{\sqrt{{3}}}{{2}}\right)}={6}+{6}{i}\sqrt{{3}}={6}{\left({1}+{i}\sqrt{{3}}\right)}$$.

### Relevant Questions

Write the standard form of the complex number.
$$\displaystyle{12}{\left({\cos{{\left({60}^{\circ}\right)}}}+{i}{\sin{{\left({60}^{\circ}\right)}}}\right)}$$
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Which statement is correct?
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}\le{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{<}{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}{>}{\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle\frac{{{3.56}\cdot{10}^{{2}}}}{{{1.09}\cdot{10}^{{4}}}}={\left({4.08}\cdot{10}^{{2}}\right)}{\left({1.95}\cdot{10}^{{-{{6}}}}\right)}$$
$$\displaystyle{9}^{{x}}-{3}^{{{x}+{1}}}+{1}={0}$$