Question

Z=z₁/z₂+z₃ z₁=-3-2i z₂=1+9i z₃=-6+i z=?

Answer 1

$\displaystyle Z= \frac{z_1}{z_2}+z_3\\\\Z= \frac{-3-2i}{1+9i}+(-6+i)= \frac{(-3-2i)(1-9i)}{(1^2-(9i)^2)}+(-6+i)=\\\\= \frac{-3-2i+27i+18i^2}{1-81i^2}+(-6+i)=\\\\= \frac{-3+25i-18}{1+81}+(-6+i)= \frac{25i-21}{82}+ \frac{82(-6+i)}{82}=\\\\= \frac{25i-21-492+82i}{82}= \frac{107i-513}{82}= \frac{107}{82}i- \frac{513}{82}$

Answer 2

To find the value of z, you need to substitute the given values of z₁, z₂, and z₃ into the expression Z = z₁/z₂ + z₃ and then perform the arithmetic.

Let's plug in the values:

z₁ = -3 - 2i z₂ = 1 + 9i z₃ = -6 + i

Z = (-3 - 2i)/(1 + 9i) + (-6 + i)

Now, to simplify the expression, we'll rationalize the denominator of the first fraction by multiplying both the numerator and denominator by the conjugate of the denominator:

(conjugate of denominator) = (1 - 9i)

Now, let's perform the rationalization:

Z = [(-3 - 2i)(1 - 9i)] / [(1 + 9i)(1 - 9i)] + (-6 + i)

Now, expand the numerator and the denominator:

Z = [-3 + 27i - 2i + 18i^2] / (1 - (9i)^2) - 6 + i

Since i^2 is equal to -1:

Z = [-3 + 27i - 2i + 18(-1)] / (1 - 81i^2) - 6 + i Z = [-3 + 27i - 2i - 18] / (1 + 81) - 6 + i

Combine like terms:

Z = (24i - 21) / 82 - 6 + i

Now, separate the real and imaginary parts:

Z = (-21/82 - 6) + (24/82)i

Finally, perform the arithmetic:

Z = -129/41 + (12/41)i

So, the value of z is:

z = -129/41 + (12/41)i