Whakaoti i te 1/1+i+i

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

Aromātai

$$\frac{1}{2}+\frac{1}{2}i=0.5+0.5i$$

Ngā Upane Otinga

Me whakarea te taurunga me te tauraro o \frac{1}{1+i} ki te haumi hiato o te tauraro, 1-i.

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

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Hei tōna tikanga, ko te i^{2} ko -1. Tātaitia te tauraro.

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

Whakareatia te 1 ki te 1-i, ka 1-i.

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

Whakawehea te 1-i ki te 2, kia riro ko \frac{1}{2}-\frac{1}{2}i.

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

Whakakotahitia ngā tau tūturu me ngā tau pōhewa i roto o \frac{1}{2}-\frac{1}{2}i me te i.

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

Tāpiri -\frac{1}{2} ki te 1.

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

Wāhi Tūturu

$$\frac{1}{2} = 0.5$$

Ngā Upane Otinga

Me whakarea te taurunga me te tauraro o \frac{1}{1+i} ki te haumi hiato o te tauraro, 1-i.

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

Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Hei tōna tikanga, ko te i^{2} ko -1. Tātaitia te tauraro.

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

Whakareatia te 1 ki te 1-i, ka 1-i.

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

Whakawehea te 1-i ki te 2, kia riro ko \frac{1}{2}-\frac{1}{2}i.

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

Whakakotahitia ngā tau tūturu me ngā tau pōhewa i roto o \frac{1}{2}-\frac{1}{2}i me te i.

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

Tāpiri -\frac{1}{2} ki te 1.

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

Ko te wāhi tūturu o \frac{1}{2}+\frac{1}{2}i ko \frac{1}{2}.

$$\frac{1}{2}$$