Senior 2 Mathematics Complex Number Functions, Senior 2 Mathematics Complex Numbers

f(x)=x^3+bx^2+cx+d

Because p(0,2), f(0)=d=2

f'(x)=3x^2+2bx+c

f(-1)=(-1)^3+b(-1)^2+c(-1)+d=-1+b-c+d=-1

i.e. b-c+d=0, b-c=-2

f'(-1)=3(-1) 2+2b(-1)+c=3-2b+c=0, i.e., 2b-c=3

The simultaneous solution is b=5, c=7

The analytical formula for f(x) is f(x)=x 3+5x 2+7x+2f'(x)=3x2+10x+7=(3x+7)(x+1)let f'(x)=0 The solution of x=-1 or x=-7 3 is easy to obtain x=-7 3 is the maximum point, and x=-1 is the minimum point.

So f(x) monotonically increases in the interval (-infinity, -7 3) and (1, + infinity) monotonically decreases in the interval (-7 3, -1).

Look at the time in order!! Thank you for your satisfaction!!

f(x)=x 3+bx 2+cx+d over p(, then:

f(0)=d=2

f(-1)=-1+b-c+2=-1

c=b+2f'(x)=3x 2+2bx+c is given by the known available equation f'(x)=3x 2+2bx+b+2=0 is -1, x+1)(3x+b+2)=0 [decomposition factor].

3+b+2=2b [the primary term is equal to the original equation].

b=5, c=b+2=7

Another: -b+2) 3=-7 3

So y=f(x)=x 3+5x 2+7x+2 from two zeros -7 3,-1 and the opening direction, we can know:

Minus interval (-7, 3, -1).

increase interval (- 7 3), 1,+

Solution: Substitute the point p(0,2) into f(x)=x 3+bx 2+cx+d, and get d=2

f(x) is the extreme value of the point x=-1, and f is obtained'(-1)=3x 2+2bx+c=0, i.e. 3-2b+c=0 (1).

f(x) points the extreme value -1 at the point x=-1, and gives f(x)=-1, i.e., -1+b-c+2=-1 (2).

1), (2) synthesis, get b = 5, c = 7

So y=f(x)=x 3+5x 2+7x+2 let f'(x)=3x 2+10x+7=0, x=-1 or x=-7 3 is easy to know x=-7 3 is the maximum point, and x=-1 is the minimum point.

So the monotonically increasing interval of f(x) is (-infinity, -7 3) and (-1, +infinity) monotonically decreasing interval is (-7 3, -1).

1. Let the real number solution be m,n

Then: 2m+1=n

1=3-n2m+an=9

n-4m-b=-8 (the i of the second equation is outside) is solved: m=1 2, n=2, a=4, b=8

2. Your question is |z|-z=2i (2+i)?

Let Z=A+Bi

Root(a2+b 2)-a-bi=(2+4i) 5So, root capacity(a2+b 2)-a=2 5

b = 4 5 solution: a = 3 5, b = -4 5

z=3/5-4/5i

<1>2x+1=y

1=3-y gives x=1 2, y=2

2x+ay=9 is equivalent to 1+2a=9

y-4x-b=-8 is equivalent to -b=-8

The solution yields a=4 and b=8

2> the topic is not clear enough, isn't "z-z" equal to 0, is it a conjugate plural?

There is a real number solution, x, y are real numbers, the real part and the imaginary part in the formula correspond equally, the first formula has 2x+1=y,,1=3-y;;; The second formula has 2x+ay+y-4x=9,b=8;;; Understandable.

Solution: a, b are the two inner angles of an acute triangle.

a+b>π/2

a>π/2-b>0, b>π/2-a>0

y=tanx is an increment function on (0, 2).

So tana>tan(2-b), tanb>tan(2-a).

i.e. tana>cotb, tanb>cota, so cotb-tana<0, tanb-cota>0, so z=(cotb-tana)+i(tanb-cota) corresponding to the point (cotb-tana, tanb-cota) is in the second quadrant.

The value of the square root of the sum of the squares of the real and imaginary parts of a complex number is called the modulo of the complex number and is denoted as z z = a+i

z∣=√(a^2+1^2)

So the range is (1, 5)c

One. Let z=a+bi, by |z-1|=1 to get |a-1+bi |=1, i.e. the square of (a-1) + the square of b = 1....1)

1 z=1 (a+bi)=(a-bi) (a-bi)(a+bi)=(a-bi) (square of a + square of b). From z+1 z as a real number, the imaginary part of z+1 z, i.e., b-b (the square of a + the square of b), must be zero. Simplification, the square of a + the square of b = 1....

2) Simultaneous (1) (2) two formulas, a = 1 2, b = root number 3 2 or b = - root number 3 2

Let Z=A+Bi, with Z+1 Z as a real number, and substituting Z into the solution to obtain a 2+b 2=1 from the latter condition (A-1) 2+b 2=1, let's calculate slowly!

That is, when 2m2-3m-2 (m2-3m+2).

Derive m 0 or, m 2

In this case, the complex number is -2+2i or 0

b, the square of a pure imaginary number is the square of the real number multiplied by the square of i, i 2=-1, so it is a negative real number.