Square root of the complex number-16-30i.
Answers
Answer:
Step-by-step explanation:
a² - b² = x = -16
2ab = y = -30
The trick is that when you square a binomial that's a difference of two terms (like a² - b²), the middle term in the expansion has a - sign; and if you can reverse that to a + sign, you'll have the square of another binomial, which is the *sum* of the same two terms you started out with the *difference* of. (Whew! What??)
OK, here's what I mean:
(a² - b²)² = a⁴ - 2a²b² + b⁴
Then to reverse the sign of the middle term, add 4a²b² to both sides:
(a² - b²)² + 4a²b² = a⁴ + 2a²b² + b⁴ = (a² + b²)²
But 4a²b² = (2ab)² = y²
and a² - b² = x
both of which you already know. So what you can do is:
(a² + b²)² = (a² - b²)² + 4a²b² = x² + y²
And now you have the sum and difference of a² and b², which makes it easy to solve for each one separately:
(a² + b²)² = x² + y² = (-30)² + (-16)² = 900 + 256 = 1156
a² + b² = √1156 = 34
a² - b² = x = -16
a² = ½(34 + [-16]) = 9; a = ±3
b² = ½(34 - [-16]) = 25; b = ±5
This gives you 4 choices of sign combination, only 2 of which will work in the original problem. So just find one that works, and the other has to be the sign-reversal of both terms (you could also figure it out by finding the quadrant the original z lies in, then look at half that z's angle):
(a + bi)² =
(3 + 5i)² = -16 + 30i . . . XX
(3 - 5i)² = -16 - 30i . . . √√
So your two answers would be
3 - 5i, -3 + 5i
Note, however, that when you ask for "the" square root, you're asking not for both solutions of
w² = z
but for the one in the agreed-upon "zone." Because square root can be taken as either root of that equation, the square root function, which must be single-valued, can *be* only one of them.
On the real line, we always specify the non-negative one.
In the complex plane, a "cut line" is defined, which usually places the argument (angle) of z either in
the range, [0,2π), or
the range, (-π,π];
which puts the angle of √z either in
[0,π) or
(-½π,½π].
If the [0,2π) range is chosen for z, then your solution is w = -3 + 5i
If the (-π,π] range is chosen for z, then your solution is w = 3 - 5i
See in attachment. .
Hope its help .u. . .
