What is e^{it}(-i+1) ?

The solution to e^{it}(-i+1) is (cos(t)+sin(t))+i(-cos(t)+sin(t))

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simplify e^{it}(-i+1)

simplify

e^{i t} (- i + 1)

(cos (t) + sin (t)) + i (- cos (t) + sin (t))

Solution steps

e^{i t} (- i + 1)

e^{i t} = cos (t) + i sin (t)

= (1 - i) (cos (t) + i sin (t))

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

a = cos (t), b = sin (t), c = 1, d = - 1

= (cos (t) \cdot 1 - sin (t) (- 1)) + (cos (t) (- 1) + sin (t) \cdot 1) i

Simplify

cos (t) \cdot 1 - sin (t) (- 1) : cos (t) + sin (t)

Simplify

cos (t) (- 1) + sin (t) \cdot 1 : - cos (t) + sin (t)

= (cos (t) + sin (t)) + (- cos (t) + sin (t)) i

Remove parentheses:

(a) = a

= cos (t) + sin (t) + (- cos (t) + sin (t)) i

Rewrite

cos (t) + sin (t) + (- cos (t) + sin (t)) i

in standard complex form:

(cos (t) + sin (t)) + (- cos (t) + sin (t)) i

= (cos (t) + sin (t)) + (- cos (t) + sin (t)) i

\frac{y ^{'} ( t )}{( m ( t ) - y ) ( y - m ( t ) )}

5 (2 x y^{2} + 2 y \frac{d}{d x} ((y) x^{2}))

r (2, 2, - 1) \cdot s (1, - 2, 3)

ln (7 x^{2})

\frac{( ( λ - 1 ) ( 4 λ - 3 ) )}{4}

What is e^{it}(-i+1) ?
The solution to e^{it}(-i+1) is (cos(t)+sin(t))+i(-cos(t)+sin(t))