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

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

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simplify (1+i)e^{it}

simplify

(1 + i) e^{i t}

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

Solution steps

(1 + i) e^{i t}

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

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

Simplify

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

Simplify

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

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

Remove parentheses:

(a) = a

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

Rewrite

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

in standard complex form:

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

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

8 (r (x) cos (x) + r (x) sin (x) = 9)

\frac{3 x}{( x + 2 ) ( x + 3 ) ( x + 1 )}

\frac{8 x ( 5 x + 7 )}{( 7 x - 5 ) ^{2}}

\frac{n ( n + 1 )}{n + 2}

\frac{1}{2 ( 4 x - 10 )}

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