Let Z satisfy | z-2-3i | = 1, and find the maximum of | Z |

Let Z satisfy | z-2-3i | = 1, and find the maximum of | Z |

1=|z-2-3i|=|z-(2+3i)|≥|z|-|2+3i |,
So | Z | ≤ 1 + | 2 + 3I | = 1 + √ 13
If the complex Z satisfies | Z-4 + 3I | ≤ 1 and the maximum value of | Z + 4-3i | is m and the minimum value is n, then m · n=_____ ?
Let z = a + bi
|z-4+3i|≤1
(a-4)^2+(b+3)^2
Given the complex number Z = | 2-3i | + 3I, find | Z|
Z = | 2-3i | + 3I = radical 13 + 3I
|Z | = radical (13 + 9) = radical 22
z=|2-3i|+3i=7+3i
|z|=49+9=58
Given the complex Z satisfies: | Z | = 1 + 3i-z, find [(1 + I) ^ 3 * (3 + 4I)] / Z
Let z = 1 - | Z | + 3I = a + 3I, then 1 - | Z | = a, | Z | ^ 2 = (1-A) ^ 2 = a ^ 2 + 9, the solution is a = - 4, the solution is 2 + 2I
Let z = 3I + X and substitute it into the equation to get x = - 4. Then substitute it into the expression to get the result
The real part of complex z = 1-3i1 + I is ()
A. 2B. -1C. 1D. -4
∵ z = 1-3i1 + I = (1-3i) (1-I) (1 + I) (1-I) = - 2-4i2 = - 1-2i ∵ the real part of the complex number is - 1, so choose B
The point corresponding to the third power of (1-3i) is ()
A first quadrant B second quadrant C third quadrant D fourth quadrant
Can give the correct answer directly
?
Don't worry if you choose D 3 power, because the sign doesn't change, as long as the denominator inside is real
What is the 9th power of the complex number (2 - √ 3I)?
Hello
(2-√3i)＾3
=2＾3-3*2＾2*√3i+3*2*（-3）+3√3i
=-10-9√3i
(2-√3i)＾9
=（-10-9√3i）＾3
=-[1000+300*9√3i+30*（-243） -243*9√3i）
=-（-6290+513√3i）
=6290-513√3i
It turns out to be such a strange number
no way out
If you don't understand this question, you can ask. If you are satisfied, remember to adopt it
I wish you progress in your study!
It is known that O is the origin of coordinates, and vectors OZ1 and oz2 correspond to complex numbers Z1 and Z2 respectively
Let o be the origin of coordinates, and the vectors OZ1 and oz2 correspond to the complex numbers Z1 and Z2 respectively,
And Z1 = 3 / (a + 5) + (a ^ 2-10) I, Z2 = 2 / (1-A) + (2a-5) I, (a belongs to R),
Z1 + Z2 are real numbers
1. Find the real number a
2. Find the area of parallelogram with OZ1 and oz2 as adjacent sides
First question a = 3
Help to do the second question directly
From the first question, a = 3, so Z1 (3 / 8, - 1) Z2 (- 1,1) plus the midpoint B (- 5 / 16,0) of origin (0,0) z1z2, it is easy to find another vertex is C (- 5 / 8,0), so the area is 2 * 1 * 5 / 8 = 5 / 4
In the complex plane, 0 is the origin, and the complex number corresponding to the vector AB is 2 + I. if point a is symmetric to point B about the real number axis, find the complex number corresponding to the vector ob
Let's draw a complex plane. Point a is symmetric about the real axis, and point B is symmetric about the X axis. Vector ob = 2-i
The conjugate complex number of 5 + 5I / 2-I is
De denominator
It can be reduced to 1 + 3I
The conjugate complex number is 1-3i