Senior 2 Mathematics Complex Numbers Please answer in detail, thank you 13 20 38 31

Knowing the complex number z=x+yi, and z-2 = 3, the point z(x,y) satisfies (x-2) +y =3, and the slope of the line connecting the point and the origin on the circumference.

y x=k has a maximum value of 3

So the maximum value of y x is 3.

z=x+yi, and z-2 = 3

then x-2+yi = 3

Let y x=k

then x-2+kxi = 3

So, (x-2) 2+(kx) 2=3

Simplify (1+k2)x 2-4x+1=0

When x=0, the above equation will not hold, so x is not equal to 0

So 1+k 2=(4x-1) x 2=4(1 x)-(1 x) 2=-(1 x-2) 2+4

1 x-2) 2+4 is less than or equal to 4

So 1+k 2 is less than or equal to 4

So k 2 is less than or equal to 3

So k is up to 3

The complex number z=x+yi, and z-2 = 3

x-2+iy∣=√3

x-2)²+y²=3

Let y x=k

Get (1+k) x -4x+1=0

16-4(1+k²)≥0

k²≤3k≤√3

The maximum value of y x is 3

According to (x-2) 2+y 2=3

Draw a circle with the maximum slope of the line between the point on the circle and the origin.

As can be seen from the image:

The answer is: root number 3

This is probably not a chemical ......I'm a little dizzy, so I'll tell you about it as I understand it.

From the problem, the point corresponding to the complex number (z2,-i) is on the negative half axis of the real axis, that is, the imaginary part of the complex number (z2,-i) is 0.

Therefore, with potatoes, there is m2-3m+3-1=0, and the solution is m=1 or 2 (m2 is the square of m, right?). ), then the value of the complex number z2 is -2+i or i (the one who repeats the quarrel with z1 is rounded). So m=1, z2=-2+i.

There is also z1·z=z2, let z1 be (1,a), (a r), z be (x,y), and from above, z2 is (-2,1).

So there is x=-2, y=1 aBecause a touches r, 1 a r, so plural z

The trajectory of the corresponding point is x=-2

x2-(4+i)x+k+2i=0

x2-4x+k+i(2-x)=0

x=2 and k r

k=4a=2^(log2(3-2x))=2^[(log2(3-2x))/2]=2^(log2√(3-2x))=3-2x)

x-1+xi≤√（2）^log2(3-2x)|=x-1+ix≤√(3-2x)|

It seems that the symbol of the absolute value should be enclosed before and after x-1+ix, indicating the modulo of this complex number.

So the question is:

(x-1)2+x2)≤√3-2x)

Both sides are squared at the same time,:

2x2-2x+1≤3-2x

x2≤1b=

x^2-(4+i)x+k+2i=0

x^2-4x+k+(2-x)i=0

x, k r

2-x=0x^2-4x+k=0

x=2k=4

What does that mean? Complex numbers can't compare sizes... Only the mold can...

First of all, my plural is terrible, and I can only use a very bad method.

1.Let z=sin +cos *i, then hail hunger is wanton and unbridled.

z 7 + z -1 = 0, so sin7 +cos7 *i+sin +cos *i-1=0, so.

sin7θ+sinθ=1，cos7θ+cosθ=0.

So sin4 cos3 = 1 grandson return 2 and cos4 cos3 =0

So cos4 = 0, cos3 = +1 2, sin4 =+1 and then later, hehe, let's discuss.

2.Let m=a+b*i, divide the real part into the imaginary part, then a=f(x), b=g(x), |m|=(a^2+b^2)^(1/2)=(x+1/x)^2+（2/x）^2)^1/2

min is obviously obtained at x=+-5) (1 2) and is (2+2(5) (1 2)) 1 2).

Why is it such an ugly answer, just talk about ideas, don't blame if it's wrong.

z1*z2=2+2i+xi-x=2-x+(2+x)i>0.Then 2+x=0, that is, z1*z2 is a real number to change to cautious sun, and imaginary numbers cannot compare the size of the nucleus, so x=-2

The cube root of 1 is (cos2 3 isin (cos4 3 isin4 3).

Then: the cube root of 1 8 is: 1 2, (1 2) (cos2 3 isin2 3), (1 2) (cos4 3 isin4 3).

The cubic roots of 27 are: 3, 3 (cos2 3 isin2 3), 3 (cos4 3 isin4 3).

There are 3 cube roots of 1, which are -1 2+isqr(3) 2, -1 2-isqr(3) 2, -1

The cube root of 1 8 is equal to the cube root of 1* (1 2).

The cube root of 27 is equal to the cube root of 1 * (-3).

We know that the modulo of a vector is equal to the square of the vector under the root sign. z-z1 = root number ((z-z1) 2) = root number (z 2 + z1 2-2 * z * z1 *cos), when cos = -1, the value is the maximum, z + z1 = 2 root number 2 + 1.