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 agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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