The set of complex Z satisfying the conditions | Z | = 1 and | Z + 1 / 2 | = | Z-3 / 2 |, is

The set of complex Z satisfying the conditions | Z | = 1 and | Z + 1 / 2 | = | Z-3 / 2 |, is

Answer: a set of two points: {z = 1 / 2 + I radical 3 / 2, 1 / 2 - I radical 3 / 2}
Analysis: or let the complex number be z = x + Yi, and the first condition becomes:
X ^ 2 + y ^ 2 = 1, (^ 2 is the square), which is a unit circle;
The second condition is equivalent to:
(x + 1 / 2) ^ 2 + y ^ 2 = (x - 3 / 2) ^ 2 + y ^ 2
X = 1 / 2, which is a straight line parallel to the y-axis
Y = 3 / 2, so the set is
Z = 1 / 2 + I radical 3 / 2, 1 / 2 - I radical 3 / 2 (I is an imaginary unit)
If you have any questions, please continue to ask~
When the complex Z satisfies what conditions, Z / (1 + Z ^ 2) belongs to R
Given that I is an imaginary unit, α is a real number, z = sin α + 1 / I to the power of 2013, and the conjugate complex of Z is Z1, then the value range of Z · Z1 is
Specific process
Z=(sinα+1/i)^2013=(sinα-i)^2013
Z1=(sinα+i)^2013
ZZ1=(sinα-i)^2013(sinα+i)^2013=[(sinα+i)(sinα+i)]^2013=(sin²α+1)^2013
∵1≤sin²α+1≤2
∴1≤ZZ1≤2^2/13
Let the complex number Z1 ≠ 1, (z1-1) / (z1 + 1) be a pure imaginary number, and find the trajectory equation of the point corresponding to the complex number Z = 4 / (1 + z1) ^ 2
Let Z1 = a + bi, where a and B are real numbers. Then (z1-1) / (z1 + 1) = [(A-1) + bi] / [(a + 1) + bi] = [(A & sup2; - 1 + B & sup2;) + (2b) I] / [(a + 1) & sup2; + B & sup2;] are pure imaginary numbers, then a & sup2; + B & sup2; = 1 and B ≠ 0. Let z = x + Yi, then x + Yi = 4 / [(a + 1) + bi] & sup2; = 4
Let the complex Z satisfy 4Z + 2Z (there is a cross on Z) = 3 √ 3 + I, and find the modulus of the complex Z
Let z = a + bi, because 4Z + 2Z (Z has a cross) = 3 √ 3 + I, so 4 (a + bi) + 2 (a-bi) = 3 √ 3 + I, that is, 6a + 2bi = 3 √ 3 + I, 6a = 3 √ 3, 2b = 1, a = √ 3 / 2, B = 1 / 2, so the modulus of complex Z is 1
How can the complex number 5 / 1 + I ^ 3 become a + bi!
LZ, your formula is obviously wrong, 5 / 1 = 5. If it is like this, it will be directly 5-I, directly in the form of a + bi. It's too easy. The correct formula should be 5 / I + I ^ 3. The front 5 / I, the top and bottom are multiplied by I, that is - 5i-i, and the final answer is - 6I. The form of a + bi is 0 - (- 6) I
LZ means 5 / (1 + I ^ 3)
Original formula = 5 / (1-I) numerator and denominator multiply by (I + 1)
The original formula is 2 + 5
So far, it has become the standard type
Is different kinds of singular after different kinds of and plural after different kinds of?
There is no difference kind of
Only different kinds of, followed by singular, plural and uncountable nouns
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Let the complex Z satisfy 4Z + 2 (Z -) = 3 (√ 3) + 1, and find the modulus of the complex Z
A + bi, a and B are real numbers
Then the module | a + bi | = √ (A & sup2; + B & sup2;)
If the complex number 1-I / (1 + I) ^ 2 = a + bi (a, B is R), then B =?
(1+i)^2=1^2+2i+i^2=2i
(1-i)/2i=(1-i)i/2i^2=(i+1)/-2=-1/2-1/2i
a=b=-1/2
Do you add the plural of the noun after "kind of"?
Plural and uncountable nouns