Whakaoti i te (z+i)(z-3i)=z(z-i)

$$( z + i ) ( z - 3 i ) = z ( z - i )$$

Whakaoti mō z

$$z=-3i$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakamahia te āhuatanga tuaritanga hei whakarea te z+i ki te z-3i ka whakakotahi i ngā kupu rite.

$$z^{2}-2iz+3=z\left(z-i\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te z ki te z-i.

$$z^{2}-2iz+3=z^{2}-iz$$

Tangohia te z^{2} mai i ngā taha e rua.

$$z^{2}-2iz+3-z^{2}=-iz$$

Pahekotia te z^{2} me -z^{2}, ka 0.

$$-2iz+3=-iz$$

Tangohia te -iz mai i ngā taha e rua.

$$-2iz+3-\left(-iz\right)=0$$

Pahekotia te -2iz me iz, ka -iz.

$$-iz+3=0$$

Tangohia te 3 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$-iz=-3$$

Whakawehea ngā taha e rua ki te -i.

$$z=\frac{-3}{-i}$$

Me whakarea tahi te taurunga me te tauraro o \frac{-3}{-i} ki te wae pohewa i.

$$z=\frac{-3i}{1}$$

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

$$z=-3i$$