3 years ago

Solve the following system of equations for x using elimination

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

Step-by-step explanation:

Radda [10]3 years ago

3 0

Answer:

b 1 ksoeoeskmx

Step-by-step explanation:

ksowoxk374

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g 7. Find Re f and Im f and find their values at the given z. (Both answers should be included) f = z⁄(z + 1), z = 4 − 5

Schach [20]

Answer:

The real and imaginary parts of the result are $\frac{1441}{1601}$ and $\frac{4}{1601}$ , respectively.

Step-by-step explanation:

Let be $f(z) = \frac{z}{z+1}$ , the following expression is expanded by algebraic means:

$f(z) = \frac{z\cdot (z-1)}{(z+1)\cdot (z-1)}$

$f(z) = \frac{z^{2}-z}{z^{2}-1}$

$f(z) = \frac{z^{2}}{z^{2}-1}-\frac{z}{z^{2}-1}$

If $z = 4 - i5$ , then:

$z^{2} = (4-i5)\cdot (4-i5)$

$z^{2} = 16-i20-i20-(-1)\cdot (25)$

$z^{2} = 41 - i40$

Then, the variable is substituted in the equation and simplified:

$f(z) = \frac{41-i40}{41-i39} -\frac{4-i5}{41-i39}$

$f(z) = \frac{37-i35}{41-i39}$

$f(z) = \frac{(37-i35)\cdot (41+i39)}{(41-i39)\cdot (41+i39)}$

$f(z) = \frac{1517-i1435+i1443+1365}{3202}$

$f(z) = \frac{2882+i8}{3202}$

$f(z) = \frac{1441}{1601} + i\frac{4}{1601}$

The real and imaginary parts of the result are $\frac{1441}{1601}$ and $\frac{4}{1601}$ , respectively.

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