What is 4-(-2+(4i)/n)^2 ?

The solution to 4-(-2+(4i)/n)^2 is (16}{n^2}+\frac{16)/n i

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simplify 4-(-2+(4i)/n)^2

simplify

4 - (- 2 + \frac{4 i}{n})^{2}

\frac{16}{n ^{2}} + \frac{16}{n} i

Solution steps

4 - (- 2 + \frac{4 i}{n})^{2}

Expand

(- 2 + \frac{4 i}{n})^{2} : \frac{4 n ^{2} - 16 in - 16}{n ^{2}}

= 4 - \frac{4 n ^{2} - 16 in - 16}{n ^{2}}

Join

4 - \frac{4 n ^{2} - 16 in - 16}{n ^{2}} : \frac{16 in + 16}{n ^{2}}

= \frac{16 in + 16}{n ^{2}}

Rewrite

16 in + 16

in standard complex form:

16 + 16 ni

= \frac{16 + 16 ni}{n ^{2}}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{16}{n ^{2}} + \frac{16 n}{n ^{2}} i

Cancel

\frac{16 n}{n ^{2}} : \frac{16}{n}

= \frac{16}{n ^{2}} + \frac{16}{n} i

n^{2} - 5 n - 2 = 0

- \frac{x}{2} \leq 1

∣ 10 - 8 x ∣ < - 13

8 y^{2} - 6 y - 9 = 0

2 x (x + 5) = - 6

What is 4-(-2+(4i)/n)^2 ?
The solution to 4-(-2+(4i)/n)^2 is (16}{n^2}+\frac{16)/n i