It's urgent If the complex Z satisfies | Z | = | Z + 2 + 2I |, find the minimum value of | Z-1 + 2I | It's best to consider the geometric meaning

It's urgent If the complex Z satisfies | Z | = | Z + 2 + 2I |, find the minimum value of | Z-1 + 2I | It's best to consider the geometric meaning

Geometric method:
|Z | = | Z + 2 + 2I | means that the distance from Z to the origin is equal to the distance from Z to - 2-2i, so the minimum value of | Z-1 + 2I | on the middle vertical line of the two points is the length of the vertical section of the line through 1-2i
Please improve the specific process yourself

It is known that the square of complex number (1 + I) to the third power × (A-I) / the square of root sign 2 × (a + 3I) is 2 / 3, so it is urgent to find a Please help me!

|[(1 + I) ^ 3 · (a - I) ^ 2] / [√ 2 (a + 3I) ^ 2] | = 2 / 32 | (a - I) ^ 2 / (a + 3I) ^ 2 | = 2 / 3 | (a - I) ^ 2 / (a + 3I) ^ 2 | = 1 / 3 | (a ^ 2 - 1 - 2ai) / (a ^ 2 + 6Ai - 9) | = 1 / 3 when (a ^ 2 - 1 - 2ai) / (a ^ 2 + 6Ai - 9) < 0 -

differentiation formulas

1. Y = C (C is constant) y '= 0
2.y=x^n y'=nx^(n-1)
3.y=a^x y'=a^xlna
y=e^x y'=e^x
4.y=logax y'=logae/x
y=lnx y'=1/x
5.y=sinx y'=cosx
6.y=cosx y'=-sinx
7.y=tanx y'=1/cos^2x
8.y=cotx y'=-1/sin^2x
9.y=arcsinx y'=1/√1-x^2
10.y=arccosx y'=-1/√1-x^2
11.y=arctanx y'=1/1+x^2
12. Y = arccotx y '= - 1 / 1 + x ^ 2, please adopt!

The formula of derivative
Who can tell me the formula of basic derivative?

(v^n)\'=nv^(n-1)(ln v)\'=v\'/v(e^v)\'=e^v*v\'(sin v)\'=cos v*v\'(cos v)\'=-sin v*v\'(tan v)\'=(sec v)^2*v\'(cot v)\'=-(csc v)^2*v\'(sec v)\'=sec v*tan v*v\'(csc v)\'=-csc v*cot v*v\'(arcsin v)\'=v\'/(...

The formula of derivation

Common derivative formula
1. Y = C (C is constant) y '= 0
　　2.y=x^n y'=nx^(n-1)
　　3.y=a^x y'=a^xlna
　　y=e^x y'=e^x
　　4.y=logax y'=logae/x
　　y=lnx y'=1/x
　　5.y=sinx y'=cosx
　　6.y=cosx y'=-sinx
　　7.y=tanx y'=1/cos^2x
　　8.y=cotx y'=-1/sin^2x
　　9.y=arcsinx y'=1/√1-x^2
　　10.y=arccosx y'=-1/√1-x^2
　　11.y=arctanx y'=1/1+x^2
　　12.y=arccotx y'=-1/1+x^2
In the process of derivation, there are several common formulas to be used:
1. In y = f [g (x)], y '= f' [g (x)] & _; G '(x)' f '[g (x)], G (x) is regarded as the whole variable, while in G' (x), X is regarded as the variable
　　2.y=u/v,y'=u'v-uv'/v^2
3. If the inverse function of y = f (x) is x = g (y), then y '= 1 / X'
Proof: 1. It is obvious that y = C is a straight line parallel to X axis, so the tangent everywhere is parallel to x, so the slope is 0. It is the same with the definition of derivative: y = C, ⊿ y = C-C = 0, Lim ⊿ x → 0 ⊿ Y / ⊿ x = 0
2. This derivation is not proved for the moment, because if it is deduced according to the definition of derivative, it can not be extended to the general case where n is any real number. After obtaining the two results y = e ^ x, y '= e ^ X and y = LNX, y' = 1 / x, it can be proved by the derivation of composite function
　　3.y=a^x,
　　⊿y=a^(x+⊿x)-a^x=a^x(a^⊿x-1)
　　⊿y/⊿x=a^x(a^⊿x-1)/⊿x
If we let ⊿ x → 0 directly, we can't derive the derivative function. We must set an auxiliary function β = a ^ ⊿ X-1 and calculate it by substitution. From the auxiliary function we can know: ⊿ x = loga (1 + β)
So (a ^ ⊿ x-1) / ⊿ x = β / loga (1 + β) = 1 / loga (1 + β) ^ 1 / β
Obviously, when ⊿ x → 0, β also tends to 0, and lim β → 0 (1 + β) ^ 1 / β = e, so Lim β → 01 / loga (1 + β) ^ 1 / β = 1 / logae = LNA
Substituting this result into Lim ⊿ x → 0 ⊿ Y / ⊿ x = Lim ⊿ x → 0A ^ x (a ⊿ x-1) / ⊿ x, we get Lim ⊿ x → 0 ⊿ Y / ⊿ x = a ^ xlna
We can know that when a = e, y = e ^ x, y '= e ^ X
　　4.y=logax
　　⊿y=loga(x+⊿x)-logax=loga(x+⊿x)/x=loga[(1+⊿x/x)^x]/x
　　⊿y/⊿x=loga[(1+⊿x/x)^(x/⊿x)]/x
Because when ⊿ x → 0, ⊿ X / X tends to 0 and X / ⊿ x tends to ∞, Lim ⊿ x → 0 log a (1 + ⊿ X / x) ^ (x / ⊿ x) = log AE, so there is
　　lim⊿x→0⊿y/⊿x＝logae/x.
We can see that when a = e, y = LNX, y '= 1 / X
In this case, we can deduce y = x ^ n, y '= NX ^ (n-1). Because y = x ^ n, y = e ^ ln (x ^ n) = e ^ nlnx,
So y '= e ^ nlnx &; (nlnx)' = x ^ n &; n / x = NX ^ (n-1)
　　5.y=sinx
　　⊿y=sin(x+⊿x)-sinx=2cos(x+⊿x/2)sin(⊿x/2)
　　⊿y/⊿x=2cos(x+⊿x/2)sin(⊿x/2)/⊿x=cos(x+⊿x/2)sin(⊿x/2)/(⊿x/2)
So Lim ⊿ x → 0 ⊿ Y / ⊿ x = Lim ⊿ x → 0cos (x + ⊿ X / 2) & _; Lim ⊿ x → 0sin (⊿ X / 2) / (⊿ X / 2) = cosx
6. Similarly, we can derive y = cosx, y '= - SiNx
　　7.y=tanx=sinx/cosx
　　y'=[(sinx)'cosx-sinx(cos)']/cos^2x=(cos^2x+sin^2x)/cos^2x=1/cos^2x
　　8.y=cotx=cosx/sinx
　　y'=[(cosx)'sinx-cosx(sinx)']/sin^2x=-1/sin^2x
　　9.y=arcsinx
　　x=siny
　　x'=cosy
　　y'=1/x'=1/cosy=1/√1-sin^2y=1/√1-x^2
　　10.y=arccosx
　　x=cosy
　　x'=-siny
　　y'=1/x'=-1/siny=-1/√1-cos^2y=-1/√1-x^2
　　11.y=arctanx
　　x=tany
　　x'=1/cos^2y
　　y'=1/x'=cos^2y=1/sec^2y=1/1+tan^2x=1/1+x^2
　　12.y=arccotx
　　x=coty
　　x'=-1/sin^2y
　　y'=1/x'=-sin^2y=-1/csc^2y=-1/1+cot^2y=-1/1+x^2
In addition, in the derivation of hyperbolic functions such as SHX, CHX, thx, and anti hyperbolic functions such as arshx, archx, arthx, and other complex composite functions, we use the derivative table and the formula at the beginning
4. Y = u soil V, y '= u' soil V '
　　5.y=uv,y=u'v+uv'
The results can be obtained quickly

Derivation of all formulas?

Methods of derivation (1) steps of deriving function y = f (x) at x0: ① increment of function Δ y = f (x0 + Δ x) - f (x0); ② average rate of change; ③ limit to get derivative. (2) derivative formulas of several common functions: ① C '= 0 (C is constant); ② (x ^ n)' = NX ^ (n-1) (n ∈ q); ③ (SiNx) '

How to find the limit of logarithm?

1. If it is the logarithm of a constant, there is no limit, because it has no process of change;
In other words, the limit is itself
2. If it is the logarithm of a function, the method to find the limit is as follows:
A. Four basic formulas of logarithm;
B. Two basic limits, or Equivalent Infinitesimal Substitution;
C. The simplification of algebra;
D. Simplification of trigonometric functions
E. The law of rotunda

In the process of using logarithm to find limit, the problem of removing logarithm sign is discussed
For example, find the limit Lim x (LN (x + 1) - ln (x + 2)) when x tends to be positive infinity
The process is like this: Lim x * ln [(x + 1) / (x + 2)] = Lim x * [(x + 1) / (x + 2) - 1] I want to know how this step changes

Ln (x + 1) / (x + 2) = ln (1 + (x + 1) / (x + 2) - 1). The latter term tends to 0 as X tends to infinity, so it can be replaced by equivalent infinitesimal

Limit logarithm
Find LIM (x → infinity) (- 2x ^ 2 * ln (x ^ 2 + 2 / x ^ 2 + 1))

x→∞,1/(x^2+1)→0,
——》ln[(x^2+2)/(x^2+1)]=ln[1+1/(x^2+1)]~1/(x^2+1),
——》Original formula = limx →∞ - 2x ^ 2 / (x ^ 2 + 1)
=limx→∞ -2/(1+1/x^2)
=-2/(1+0)
=-2.

Logarithm and limit
Find ln (2x + 1) / ln2x, where n - > infinity

0
ln(2x+1)/ln2x
=ln[(2x+1)-2x]
= ln 1
=0