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

The solution to 2(-2+(3i)/n)^2 is (8-(18}{n^2})-\frac{24)/n i

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simplify 2(-2+(3i)/n)^2

simplify

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

(8 - \frac{18}{n ^{2}}) - \frac{24}{n} i

Solution steps

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

Expand

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

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

Apply the distributive law:

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

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

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

Simplify

2 (- 2)^{2} + 2 \cdot 2 (- 2) \frac{3 i}{n} + 2 (\frac{3 i}{n})^{2} : 8 - \frac{18}{n ^{2}} - \frac{24}{n} i

= 8 - \frac{18}{n ^{2}} - \frac{24}{n} i

Rewrite in standard complex form:

(8 - \frac{18}{n ^{2}}) - \frac{24}{n} i

= (8 - \frac{18}{n ^{2}}) - \frac{24}{n} i

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

- 5 (2 - 3)

(- 3 + 5 i, - 3 + 3 i, 10 - 3 i) \cdot (- 3 - 2 i, - 4 - 3 i, 4 + 2 i)

x_{1} (x^{4} = 2 x + 1)

\frac{1}{x ^{\frac{3}{4}} ( x + 1 - x )}

What is 2(-2+(3i)/n)^2 ?
The solution to 2(-2+(3i)/n)^2 is (8-(18}{n^2})-\frac{24)/n i