What is expand 3(1+2/n i)^2 ?

The solution to expand 3(1+2/n i)^2 is (3-(12}{n^2})+\frac{12)/n i

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expand 3(1+2/n i)^2

expand

3 (1 + \frac{2}{n} i)^{2}

(3 - \frac{12}{n ^{2}}) + \frac{12}{n} i

Solution steps

3 (1 + \frac{2}{n} i)^{2}

Expand

(1 + \frac{2}{n} i)^{2} : 1^{2} + 2 \cdot 1 \cdot \frac{2}{n} i + (\frac{2}{n} i)^{2}

= 3 (1^{2} + 2 \cdot 1 \cdot \frac{2}{n} i + (\frac{2}{n} i)^{2})

Apply rule:

a \cdot 1 = a

= 3 (1^{2} + 2 \cdot \frac{2}{n} i + (\frac{2}{n} i)^{2})

Apply the distributive law:

m (a + b + c) = ma + mb + m c

3 (1^{2} + 2 \cdot \frac{2}{n} i + (\frac{2}{n} i)^{2}) = 3 \cdot 1^{2} + 3 \cdot 2 \cdot \frac{2}{n} i + 3 (\frac{2}{n} i)^{2}

= 3 \cdot 1^{2} + 3 \cdot 2 \cdot \frac{2}{n} i + 3 (\frac{2}{n} i)^{2}

Simplify

3 \cdot 1^{2} + 3 \cdot 2 \cdot \frac{2}{n} i + 3 (\frac{2}{n} i)^{2} : 3 + \frac{12}{n} i - \frac{12}{n ^{2}}

= 3 + \frac{12}{n} i - \frac{12}{n ^{2}}

Rewrite in standard complex form:

(3 - \frac{12}{n ^{2}}) + \frac{12}{n} i

= (3 - \frac{12}{n ^{2}}) + \frac{12}{n} i

(n + 7) lo g_{7} (x)

(\frac{x ^{3}}{2} + 4 x^{2} - \frac{7 x}{9}) (2 x + 5)

- \frac{2}{s ^{3} ( s ^{2} + 1 )}

\frac{cot ( x )}{sec ( x )}

\frac{( s ^{2} + 1 )}{( s + 1 ) ^{3}}

What is expand 3(1+2/n i)^2 ?
The solution to expand 3(1+2/n i)^2 is (3-(12}{n^2})+\frac{12)/n i