Let’s assume z = (x – iy) (3 + 5i)

z = 3x + 5xi - 3yi - 5yi^{2} = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)

\( \overline{z}\) = (3x + 5y) - i(5x - 3y)

It is given that, \( \overline{z}= -6 - 24i\)

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

On equating real and imaginary parts, we have

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

Performing (i) x 3 + (ii) x 5, we get

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = \( \dfrac{102}{34}\) = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3 respectively.

Answered by Pragya Singh | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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