Let z1 = 2 – i, z2 = -2 + i. Find

$$Re(\dfrac{z_1z_2}{z_1})$$  $$Im(\dfrac{1}{z_1\overline{z_1}})$$

Asked by Abhisek | 1 year ago |  106

##### Solution :-

Complex numbers

z = 2 - i,z = -2 + i

z z = (2 - i)(-2 + i) = -4 + 2i + 2i - i

= -4 + 4i - (-1) = -3 + 4i

z = 2 + i

$$\dfrac{z_1z_2}{z_1}=\dfrac{-3+4i}{2+i}$$

On multiplying numerator and denominator by (2-i) , we obtain

$$\dfrac{z_1z_2}{z_1}=\dfrac{(-3+4i)(2-i)}{(2+i)(2-i)}$$

$$\dfrac{-6+3i+8i-4i^2}{2^2+1^2}$$

$$\dfrac{-6+11i-4(-1)}{2^2+1^2}$$

$$\dfrac{-2+11i}{5}$$

$$\dfrac{-2}{5}+ \dfrac{11}{5}i$$

On comparing real parts, we obtain

$$Re(\dfrac{z_1z_2}{z_1})=\dfrac{-2}{5}$$

$$\dfrac{1}{z_1\overline{z_1}}=\dfrac{1}{(2-i)(2+i)}$$

$$=\dfrac{1}{(2)^2+(1)^2}=\dfrac{1}{5}$$

On comparing imaginary parts, we obtain

$$Im( \dfrac{1}{z_1\overline{z_1}})=0$$

Hence, solved

Answered by Abhisek | 1 year ago

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