Let’s assume z = (x – iy) (3 + 5i)
z = 3x + 5xi - 3yi - 5yi2 = 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 + i10 + i20 + i30 is a real number?
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