Ask a plural question and help solve it

Typing it wrong, right? It's 8+6i, right? Let z = a + bi, then the square of z = (the square of a - the square of b) + 2abi so (the square of a a - the square of b) = 8, 2ab = 6 The solution gives a = 3, b = 1 or a = -3, b = -1

Let Z1=A+Bi and Z2=C+Di

z1-2+i|^2=|a-2+(b+1)i |^2=(a-2)^2+(b+1)^2=1

Z2+(1+I)Z1=C+Di+(1+I)(A+Bi)=C+A-B+(D+B+A)I=0, i.e., C+A-B=0, (D+B+A)=0

From (a-2) 2+(b+1) 2=1 , c=b-a, d=-b-a:

Point A is on a circle with a center of (2,-1) and a radius of 1, and then you can combine the numbers and shapes to analyze the point A on the circle and the point outside the circle.

b(b-a,-b-a) and the position relationship of the origin, I won't analyze the specifics, you can brew it yourself and see if you can do it.

z1-2+i|The geometric meaning of =1 a is a point on a circle with (2,-1) as the center and 1 as the radius.

Let a(x,y) let x=cosa+2,y=sina-1, z1=x+yi substitute z2 to get z2=y-x-(x+y)i

Point b coordinates (y-x, -x-y).

From the geometric meaning, point b changes with the change of a, and when a is at the rightmost end of the circle (3,-1) sa0b = 5 is maximum.

When a is at the leftmost end of the circle (1,-1), saob=1 is the smallest.

Let the complex number z=a+bi, then block z.

A total of simple Cong Xiang Zheng potato yoke plural.

If z'=a-bi, it depends on the title.

z*z')-z'=10/(1-2i)

10(1+2i)/[1-2i)(1+2i)]10(1+2i)/(1+2^2)

10(1+2i)/5

2(1+2i)

2+4i and z*z'=(a+bi)*(a-bi)=a 2-(bi) 2=a 2+b 2, so the equation becomes.

a 2+b 2)-(a-bi)=2+4i, (a 2+b 2)-a]+bi=2+4i, then there is.

a 2 + b 2)-a = 2, b = 4, solution.

a=3, b=4, so.

Complex number z=a+bi=3+4i.

x is not equal to 1, so multiply both sides of the equation by x-1.

x^3-1=0

x^3=1x^30 + x^40 + x^50=(x^3)^10+x*(x^3)^13+x^2*(x^3)^16

1+x+x^2=0

c is the following complex number a+b*j, if it is to be expressed as exponential form c=r(cosu+j*sinu), therefore.

c|=r=a/cosu, tgu=rcosu/(rsinu)=a/b u=arg(a/b)

So there are two formulas in the box.

You think of complex numbers as vector coordinates.

f=f1+f2=, that is, the angle of inclination of the vector f from (0,0) to (, is arctan(y x)=arctan(|f is the modulo of the vector.

There is a real solution m, then.

m^2+am+c)+(bm+d)i=0

So m2+am+c=0 and bm+d=0

(-a+ (a 2-4c)) 2=-d b or (-a- (a 2-4c)) 2=-d b

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).

z1 times z2 is a complex number, and the membrane is the root number (cos sin +1) 2 + (cos -sin) 2

Then the square of the membrane is (cos sin +1) 2 + (cos -sin) 2