simplify (3z+1-i)^2

Solution

simplify

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

(9 z^{2} + 6 z) + (- 2 - 6 z) i

Solution steps

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

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

= (3 z + 1 - i) (3 z + 1 - i)

Distribute parentheses

= 3 z \cdot 3 z + 3 z \cdot 1 + 3 z (- i) + 1 \cdot 3 z + 1 \cdot 1 + 1 \cdot (- i) - i 3 z - i 1 - i (- i)

Simplify

3 z \cdot 3 z + 3 z \cdot 1 + 3 z (- i) + 1 \cdot 3 z + 1 \cdot 1 + 1 \cdot (- i) - i 3 z - i 1 - i (- i) : - 2 i + 9 z^{2} + 6 z - 6 z i

= - 2 i + 9 z^{2} + 6 z - 6 z i

Rewrite in standard complex form:

(9 z^{2} + 6 z) + (- 2 - 6 z) i

= (9 z^{2} + 6 z) + (- 2 - 6 z) i

- \frac{d}{d x} (\frac{2}{x ( x ^{2} + 4 )})

(2 + \frac{1}{4})^{8} (2 x - \frac{3}{x})^{2}

\frac{240}{( s + 1 ) ( s + 2 ) ( s + 3 ) ( s + 4 )}

(3 x + 2 y^{2}) d x + 2 x y d y

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