Whakaoti i te (1+i)^4/(1-i)^3+(1-i)^4/(1+i)^3

$$\frac { ( 1 + i ) ^ { 4 } } { ( 1 - i ) ^ { 3 } } + \frac { ( 1 - i ) ^ { 4 } } { ( 1 + i ) ^ { 3 } }$$

Aromātai

$$2$$

Ngā Upane Otinga

Tātaihia te 1+i mā te pū o 4, kia riro ko -4.

$$\frac{-4}{\left(1-i\right)^{3}}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Tātaihia te 1-i mā te pū o 3, kia riro ko -2-2i.

$$\frac{-4}{-2-2i}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Me whakarea te taurunga me te tauraro o \frac{-4}{-2-2i} ki te haumi hiato o te tauraro, -2+2i.

$$\frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Mahia ngā whakarea i roto o \frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$\frac{8-8i}{8}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Whakawehea te 8-8i ki te 8, kia riro ko 1-i.

$$1-i+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Tātaihia te 1-i mā te pū o 4, kia riro ko -4.

$$1-i+\frac{-4}{\left(1+i\right)^{3}}$$

Tātaihia te 1+i mā te pū o 3, kia riro ko -2+2i.

$$1-i+\frac{-4}{-2+2i}$$

Me whakarea te taurunga me te tauraro o \frac{-4}{-2+2i} ki te haumi hiato o te tauraro, -2-2i.

$$1-i+\frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}$$

Mahia ngā whakarea i roto o \frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}.

$$1-i+\frac{8+8i}{8}$$

Whakawehea te 8+8i ki te 8, kia riro ko 1+i.

$$1-i+\left(1+i\right)$$

Tāpirihia te 1-i ki te 1+i, ka 2.

$$2$$

Wāhi Tūturu

$$2$$

Ngā Upane Otinga

Tātaihia te 1+i mā te pū o 4, kia riro ko -4.

$$Re(\frac{-4}{\left(1-i\right)^{3}}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Tātaihia te 1-i mā te pū o 3, kia riro ko -2-2i.

$$Re(\frac{-4}{-2-2i}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Me whakarea te taurunga me te tauraro o \frac{-4}{-2-2i} ki te haumi hiato o te tauraro, -2+2i.

$$Re(\frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Mahia ngā whakarea i roto o \frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$Re(\frac{8-8i}{8}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Whakawehea te 8-8i ki te 8, kia riro ko 1-i.

$$Re(1-i+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Tātaihia te 1-i mā te pū o 4, kia riro ko -4.

$$Re(1-i+\frac{-4}{\left(1+i\right)^{3}})$$

Tātaihia te 1+i mā te pū o 3, kia riro ko -2+2i.

$$Re(1-i+\frac{-4}{-2+2i})$$

Me whakarea te taurunga me te tauraro o \frac{-4}{-2+2i} ki te haumi hiato o te tauraro, -2-2i.

$$Re(1-i+\frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)})$$

Mahia ngā whakarea i roto o \frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}.

$$Re(1-i+\frac{8+8i}{8})$$

Whakawehea te 8+8i ki te 8, kia riro ko 1+i.

$$Re(1-i+\left(1+i\right))$$

Tāpirihia te 1-i ki te 1+i, ka 2.

$$Re(2)$$

Ko te wāhi tūturu o 2 ko 2.

$$2$$