faktorisieren (e^{i2θ}-1)/(e^{i2θ)+1}

Lösung

faktorisieren

\frac{e ^{i 2 θ} - 1}{e ^{i 2 θ} + 1}

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

Schritte zur Lösung

\frac{e ^{i 2 θ} - 1}{e ^{i 2 θ} + 1}

e^{i 2 θ} - 1 = cos (2 θ) + i sin (2 θ) - 1

= \frac{cos ( 2 θ ) + i sin ( 2 θ ) - 1}{e ^{2 i θ} + 1}

e^{i 2 θ} + 1 = cos (2 θ) + i sin (2 θ) + 1

= \frac{cos ( 2 θ ) + i sin ( 2 θ ) - 1}{cos ( 2 θ ) + i sin ( 2 θ ) + 1}

Rationalisiere

\frac{cos ( 2 θ ) + i sin ( 2 θ ) - 1}{cos ( 2 θ ) + i sin ( 2 θ ) + 1} : \frac{sin ^{2} ( 2 θ ) + 2 i sin ( 2 θ ) + ( cos ( 2 θ ) - 1 ) ( cos ( 2 θ ) + 1 )}{sin ^{2} ( 2 θ ) + ( cos ( 2 θ ) + 1 ) ^{2}}

= \frac{( cos ( 2 θ ) - 1 ) ( cos ( 2 θ ) + 1 ) + sin ^{2} ( 2 θ ) + 2 i sin ( 2 θ )}{( cos ( 2 θ ) + 1 ) ^{2} + sin ^{2} ( 2 θ )}

Schreibe

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

in der Standard komplexen Form um:

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

= \frac{cos ^{2} ( 2 θ ) - 1 + sin ^{2} ( 2 θ )}{sin ^{2} ( 2 θ ) + ( 1 + cos ( 2 θ ) ) ^{2}} + \frac{2 sin ( 2 θ )}{sin ^{2} ( 2 θ ) + ( cos ( 2 θ ) + 1 ) ^{2}} i

f a c t or 4 w^{2} - 5 w - 6

f a c t or 35 + 30 s - 45 t

(2 x - 4)^{2} = 98

4 x - 3 = 3 x - 1

x^{2} + 3 x + 2 \geq 0