What is the tan invers of 3i/-1-i

Mathematics

1 answer:

postnew [5]1 year ago

6 0

z = 3i / (-1 - i )

z = 3i / (-1 - i ) × (-1 + i ) / (-1 + i )

z = (3i × (-1 + i )) / ((-1)² - i ²)

z = (-3i + 3i ²) / ((-1)² - i ²)

z = (-3 - 3i ) / (1 - (-1))

z = (-3 - 3i ) / 2

Note that this number lies in the third quadrant of the complex plane, where both Re(z) and Im(z) are negative. But arctan only returns angles between -π/2 and π/2. So we have

arg(z) = arctan((-3/2)/(-3/2)) - π

arg(z) = arctan(1) - π

arg(z) = π/4 - π

arg(z) = -3π/4

where I'm taking arg(z) to have a range of -π < arg(z) ≤ π.

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6 0

2 years ago

Read 2 more answers

City R has a temperature of -2 degrees. City S has a temperature of -6 degrees. Write an inequality to compare the temperatures

elena-s [515]

Answer:

-2 > -6 . . . or . . . R > S

Step-by-step explanation:

If you plot the temperatures of the two cities on a number line, you see that the temperature for City R is plotted to the right of the temperature for City S. This means the temperature of City S is less than that of City R.

If your inequality relates cities:

S < R or R > S

If your inequality relates temperatures:

-6 < -2 or -2 > -6