Knowing that the complex number z satisfies z 1 and z 1, it is verified that z 1 z 2 is a real nu

By |z|=1, which can be assumed.

z=cos(a)+isin(a)

According to the plural. Power operations.

Know. z^2

cos(2a)+isin(2a)

z/(1+z^2)

z/[1+cos(2a)+isin(2a)]cos(a)+isin(a)]*1+cos(2a)-isin(2a)]

1+cos(2a))²

sin(2a)²

cos(a)+cos(a)*cos(2a)+sin(a)*sin(2a)

sin(a)+sin(a)cos(2a)-cos(a)sin(2a)i

cos(a)+cos(a-2a)

sin(a)

sin(a-2a)]

i1+cos(2a))²

sin(2a)²

2cos(a)

i1+cos(2a))²

sin(2a)²

2cos(a)

1+cos(2a))²

sin(2a)²

is a real number, and the proof is complete.

Hope it helps, if you don't know please ask, it's useful.

o( o note: z (1+z 2) can be further. Simplify. For.

cos(a)

1+cos(2a)]

z=a+bi, a, b are real numbers.

then a 2 + b 2 = 1

1 z=1 (a+bi)=(a-bi) (a2+b 2)=a-bi, so z+1 z=2a

z≠ i, so a≠0

So z+1 z≠0

So z+1 z=(z 2+1) z is a real number that is not equal to 0.

So z (1+z 2) is a real number.

z=a+bi

z＋2i=a+(b+2)i

z (1 i) = (a+bi) (1 i) = (a-b) 2 + (a+b)i 2

It's all about the real number of oranges.

b+2=0a+b=0

A=2, b=-2

z=2-2i

z is the actual number of the manuscript.

The number of holes in the imaginary part is 0 [0 times any number equals 0], that is: m-2=0

m=2 Of course, when m can be an imaginary number, there are an infinite number of solutions

z-1+i) belongs to the pure void of jujube, rotten and sleepy, and the number of leaks is Yan Xun or 0

Let z=y+xi, x, y be real numbers.

z-1+i=(y-1)+(x+1)i

y-1=0 y=1

z=1+xi

z|^2=1+x^2

Let Z=A+Bi

z|=1 and z is not equal to plus or minus i, a≠0, b≠ 1a +b =1

a²=1-b²

z/(1+z²)

a+bi)/[1+(a+bi)²]

a+bi)/(a²+1-b²+2abi)=(a+bi)/(a²+a²+2abi)

a+bi)/(2a²+2abi)

a+bi)/[2a(a+bi)]

1 2a is a real number.

Known |z|=1 and z ≠ plus or minus i, i.e., the modulo of z is 1, and the spoke angle satisfies cos ≠

0, then you might as well set z=cos +isin

Bring in to obtain: z (1+z).

The denominator of 1+Z=1+(cos +isin) 2=1+cos 2-sin 2+2isin cos = 2cos 2

2isinθcosθ

2cosθ(cosθ+isinθ)

And the molecule Z = Cos + Isin

Approximate points to get:

z/(1+z²)=1/2cosθ

Apparently a real number (the coefficient of the imaginary part i is 0).

By |z|=1, it can be assumed that z=cos(a)+isin(a).

According to the power operation of complex numbers, we can know that Z2 = cos(2a)+isin(2a).

z/(1+z^2)

z/[1+cos(2a)+isin(2a)]

cos(a)+isin(a)]*1+cos(2a)-isin(2a)] / [ 1+cos(2a))²sin(2a)²

[ 1+cos(2a))²sin(2a)²

2cos(a) +0 i ] / [ 1+cos(2a))²sin(2a)²

2cos(a) / [ 1+cos(2a))²sin(2a)²

is a real number, and the proof is complete.

Note: z (1+z 2) can be further simplified to cos(a) [1+cos(2a)].

Because |z|=1 , so z*z =1, (z represents the plural number of z's common fortune and envy yoke), so (z 2+1) is z=(z 2+z*z) z=z+z is a real number.

If the equation is z 2+(1 z), the result may not be a real number.

z 2 = 1 and 1 z = 1 or -1, and the two parts add up to 0 or 2

Plural z satisfies |z+1|=1

Therefore, the complex number z is on a circle with (-1,0) as the center and 1 as the radius, then the range of the modulus of the complex number z is: 0 |z|2 then 0 |z|²≤4

Hence 1 m 5

|z+1|=1，|z|=|z+1-1|<=|z+1|+1=2, and ,|z|=|z+1-1|>=||z+1|-1|=0

m=1+|z|<=5, and m=1+|z|>=1, the value range of the real number m is [1,5].