z_{1}*z_{2}=(1-i)(sqrt{3}+i) eching

$$z _ { 1 } \cdot z _ { 2 } = ( 1 - i ) ( \sqrt { 3 } + i )$$

z_1 uchun yechish

$$z_{1}=-\frac{\sqrt{3}\left(-1+i\right)+\left(-1-i\right)}{z_{2}}$$

$$z_{2}\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

1-i ga \sqrt{3}+i ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$z_{1}z_{2}=\left(1-i\right)\sqrt{3}+\left(1+i\right)$$

Tenglama standart shaklda.

$$z_{2}z_{1}=\sqrt{3}\left(1-i\right)+\left(1+i\right)$$

Ikki tarafini z_{2} ga bo‘ling.

$$\frac{z_{2}z_{1}}{z_{2}}=\frac{\sqrt{3}\left(1-i\right)+\left(1+i\right)}{z_{2}}$$

z_{2} ga bo'lish z_{2} ga ko'paytirishni bekor qiladi.

$$z_{1}=\frac{\sqrt{3}\left(1-i\right)+\left(1+i\right)}{z_{2}}$$

z_2 uchun yechish

$$z_{2}=-\frac{\sqrt{3}\left(-1+i\right)+\left(-1-i\right)}{z_{1}}$$

$$z_{1}\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

1-i ga \sqrt{3}+i ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$z_{1}z_{2}=\left(1-i\right)\sqrt{3}+\left(1+i\right)$$

Tenglama standart shaklda.

$$z_{1}z_{2}=\sqrt{3}\left(1-i\right)+\left(1+i\right)$$

Ikki tarafini z_{1} ga bo‘ling.

$$\frac{z_{1}z_{2}}{z_{1}}=\frac{\sqrt{3}\left(1-i\right)+\left(1+i\right)}{z_{1}}$$

z_{1} ga bo'lish z_{1} ga ko'paytirishni bekor qiladi.

$$z_{2}=\frac{\sqrt{3}\left(1-i\right)+\left(1+i\right)}{z_{1}}$$