Whakaoti i te z=4-2i/1+i

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

Whakaoti mō z

$$z=1-3i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Me whakarea te taurunga me te tauraro o \frac{4-2i}{1+i} ki te haumi hiato o te tauraro, 1-i.

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

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}.

$$z=\frac{\left(4-2i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

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

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

Me whakarea ngā tau matatini 4-2i me 1-i pēnā i te whakarea huarua.

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

Hei tōna tikanga, ko te i^{2} ko -1.

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

Mahia ngā whakarea i roto o 4\times 1+4\left(-i\right)-2i-2\left(-1\right)\left(-1\right).

$$z=\frac{4-4i-2i-2}{2}$$

Whakakotahitia ngā wāhi tūturu me ngā wāhi pōhewa ki 4-4i-2i-2.

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

Mahia ngā tāpiri i roto o 4-2+\left(-4-2\right)i.

$$z=\frac{2-6i}{2}$$

Whakawehea te 2-6i ki te 2, kia riro ko 1-3i.

$$z=1-3i$$

Tautapa z

$$z≔1-3i$$