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

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

Whakaoti mō z

$$z=\frac{\sqrt{2}}{2}+\left(-\frac{1}{2}+\frac{1}{2}i\right)\approx 0.207106781+0.5i$$

$$z=-\frac{1}{2}+\frac{1}{2}i-\frac{\sqrt{2}}{2}\approx -1.207106781+0.5i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe z ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te z+1.

$$2z\left(z+1\right)=1-i+\left(z+1\right)\times \left(2i\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2z ki te z+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te z+1 ki te 2i.

$$2z^{2}+2z=1-i+2iz+2i$$

Mahia ngā tāpiri i roto o 1-i+2i.

$$2z^{2}+2z=2iz+1+i$$

Tangohia te 2iz mai i ngā taha e rua.

$$2z^{2}+2z-2iz=1+i$$

Pahekotia te 2z me -2iz, ka \left(2-2i\right)z.

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

Tangohia te 1+i mai i ngā taha e rua.

$$2z^{2}+\left(2-2i\right)z-\left(1+i\right)=0$$

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

$$2z^{2}+\left(2-2i\right)z+\left(-1-i\right)=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 2-2i mō b, me -1-i mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 2-2i.

$$z=\frac{-2+2i±\sqrt{-8i-4\times 2\left(-1-i\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$z=\frac{-2+2i±\sqrt{-8i-8\left(-1-i\right)}}{2\times 2}$$

Whakareatia -8 ki te -1-i.

$$z=\frac{-2+2i±\sqrt{-8i+\left(8+8i\right)}}{2\times 2}$$

Tāpiri -8i ki te 8+8i.

$$z=\frac{-2+2i±\sqrt{8}}{2\times 2}$$

Tuhia te pūtakerua o te 8.

$$z=\frac{-2+2i±2\sqrt{2}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$z=\frac{-2+2i±2\sqrt{2}}{4}$$

Nā, me whakaoti te whārite z=\frac{-2+2i±2\sqrt{2}}{4} ina he tāpiri te ±. Tāpiri -2+2i ki te 2\sqrt{2}.

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

Whakawehe -2+2i+2\sqrt{2} ki te 4.

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

Nā, me whakaoti te whārite z=\frac{-2+2i±2\sqrt{2}}{4} ina he tango te ±. Tango 2\sqrt{2} mai i -2+2i.

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

Whakawehe -2+2i-2\sqrt{2} ki te 4.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe z ki -1 nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te z+1.

$$2z\left(z+1\right)=1-i+\left(z+1\right)\times \left(2i\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2z ki te z+1.

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

Whakamahia te āhuatanga tohatoha hei whakarea te z+1 ki te 2i.

$$2z^{2}+2z=1-i+2iz+2i$$

Mahia ngā tāpiri i roto o 1-i+2i.

$$2z^{2}+2z=2iz+1+i$$

Tangohia te 2iz mai i ngā taha e rua.

$$2z^{2}+2z-2iz=1+i$$

Pahekotia te 2z me -2iz, ka \left(2-2i\right)z.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe 2-2i ki te 2.

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

Whakawehe 1+i ki te 2.

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

Whakawehea te 1-i, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}-\frac{1}{2}i. Nā, tāpiria te pūrua o te \frac{1}{2}-\frac{1}{2}i ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua \frac{1}{2}-\frac{1}{2}i.

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

Tāpiri \frac{1}{2}+\frac{1}{2}i ki te -\frac{1}{2}i.

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

Tauwehea z^{2}+\left(1-i\right)z-\frac{1}{2}i. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

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

Tuhia te pūtakerua o ngā taha e rua o te whārite.

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

Whakarūnātia.

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

Me tango \frac{1}{2}-\frac{1}{2}i mai i ngā taha e rua o te whārite.

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