What is z^4+((i+2+i)/(1+i*(2+i)))^3*(-i) ?

The solution to z^4+((i+2+i)/(1+i*(2+i)))^3*(-i) is (z^4-2)+2i

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simplify z^4+((i+2+i)/(1+i(2+i)))^3(-i)

simplify

z^{4} + (\frac{i + 2 + i}{1 + i \cdot ( 2 + i )})^{3} \cdot (- i)

(z^{4} - 2) + 2 i

Solution steps

z^{4} + (\frac{i + 2 + i}{1 + i ( 2 + i )})^{3} (- i)

Rewrite

i + 2 + i

in standard complex form:

2 + (1 + 1) i

= z^{4} + (\frac{2 + ( 1 + 1 ) i}{1 + i ( 2 + i )})^{3} (- i)

Add the numbers:

1 + 1 = 2

= z^{4} + (\frac{2 + 2 i}{1 + i ( 2 + i )})^{3} (- i)

Simplify

z^{4} + (\frac{2 + 2 i}{1 + i ( 2 + i )})^{3} (- i) : z^{4} - 2 + 2 i

= z^{4} - 2 + 2 i

Rewrite in standard complex form:

(z^{4} - 2) + 2 i

= (z^{4} - 2) + 2 i

\frac{( 6 ) - 7}{( 6 ) ^{2} - 49}

125 x = 25

lo g_{8} (- 6 x) < 1

69.6 - 2.539 \cdot \frac{x}{8} = 68.39

2 x^{2} + 12

What is z^4+((i+2+i)/(1+i*(2+i)))^3*(-i) ?
The solution to z^4+((i+2+i)/(1+i*(2+i)))^3*(-i) is (z^4-2)+2i