Whakaoti i te {l}{{(z+i)}{(z-3i)}=z{(z-i)}}{j=(1+i)^2}{k=j}{l=k}{text{Solvefor}mtext{where}}{m=l}

$$\left. \begin{array} { l } { {(z + i)} {(z - 3 i)} = z {(z - i)} }\\ { j = {(1 + i)} ^ {2} }\\ { k = j }\\ { l = k }\\ { \text{Solve for } m \text{ where} } \\ { m = l } \end{array} \right.$$

Whakaoti mō z, j, k, l, m

$$m=2i$$

Ngā Upane Otinga

Whakaarohia te whārite tuatahi. 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.

$$z=-3i$$

Whakaarohia te whārite tuarua. Tātaihia te 1+i mā te pū o 2, kia riro ko 2i.

$$j=2i$$

Whakaarohia te whārite tuatoru. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$k=2i$$

Whakaarohia te whārite tuawhā. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$l=2i$$

Whakaarohia te whārite tuarima. Me kōkuhu ngā uara tāupe mōhiotia ki te whārite.

$$m=2i$$

Kua oti te pūnaha te whakatau.

$$z=-3i j=2i k=2i l=2i m=2i$$