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

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

Popular >

simplify (z+i)^2*(z-1)

simplify

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

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

Solution steps

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

Expand

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

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

Distribute parentheses

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

Simplify

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

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

Rewrite in standard complex form:

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

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

e^{- 2 x} (x^{2} + x + 1) - 0.5

\frac{d ^{2}}{d ^{2} x} (x^{3} + y^{3} = 16)

\frac{sin ( x ) + cos ^{2} ( x )}{( 1 - sin ( x ) ) ^{2}}

11.18 - (5 + y) y (\frac{( ( 5 + y ) y )}{5 + 2.83 y})^{\frac{2}{3}}

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

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