What is the quotient of exponentially equal powers of the same base (not 0)?

What is the quotient of exponentially equal powers of the same base (not 0)?

Exponentially equal to the power of the same base (not 0), the quotient is 1
The same base (not 0) power division, exponential subtraction, equal to the same base 0 power, that is equal to 1.
The power of the same base with the same exponent (not 0) is equal to 1
That is, the index and the base are the same
Complex z = a + bi when a equals what B equals what Z equals zero
Title, such as
a=0; b=0
Given the complex number Z = 2 + bi and 3|z| = |z module | + 6, find the real number B and complex number Z
3 | Z | = | Z module | + 6
2|z|=6
|z|=3
therefore
4+b²=9
b²=5
b=±√5
Namely
z=2±√5i
The complex z = (1 + I) / I ^ 3|=______
∵i³=-i
∴z=(1+i)/i³=(1+i)/-i=i(1+i)=i-1
∴|z|²=1²+(-1)²=2
∴|z|=√2
If a nonzero complex Z has an angle of - 7 π / 4, why do you choose this option?
A unique angle
The principal value of B angle is unique
The main value of C is - 7 π / 4
The main value of D is 7 π / 4
1. The main value of argument is unique
2. The main value range of amplitude is [0,2 π]
So choose B
On the relationship between the sum of modulo and the divisor of complex number
Let Z1 = R1 (COSA + isina), Z2 = R1 (CoSb + isinb) (| Z1 | = R1, | Z2 | = R2, Z1 argument a, Z2 argument b),
Then Z1 / Z2 = R1 (COSA + isina) / [R1 (CoSb + isinb)] = (R1 / r2) (COSA + isina) / (CoSb + isinb)
=(r1/r2)(cosa+isina)(cosb-isinb)/[(cosb+isinb)(cosb-isinb)]
=(r1/r2)[(cosacosb+sinasinb)+(sinacosb-cosasinb)i]/[(cosb)^2+(sinb)^2]
=(R1 / r2) [cos (a-b) + isin (a-b)], the radiation angle of Z1 / Z2 is a-b,
|z1/z2|=|(r1/r2)[cos(a-b)+isin(a-b)]|=|r1/r2|√{[cos(a-b)]^2+[sin(a-b)]^2},
=|r1/r2|=|z1|/|z2|,
The module of two complex quotients = the quotient of the module, the argument of two complex quotients = the argument of the divisor - the argument of the divisor
(a+bi)/(c+di)=(a+bi)(c-di)/[(c+di)(c-di)]
=[(ac+bd)+(bc-ad)i]/(c^2-d^2i^2)
=[(ac+bd)+(bc-ad)i]/(c^2+d^2)
Consider (AC + BD) as a vector multiplied by B vector, and C ^ 2 + D ^ 2 as the module of B vector
(AC + BD) / (C ^ 2 + D ^ 2) = 1 is a * B / (b module)
But I don't want to expand the a-model either
(a+bi)/(c+di)=(a+bi)(c-di)/[(c+di)(c-di)]
=[(ac+bd)+(bc-ad)i]/(c^2-d^2i^2)
=[(ac+bd)+(bc-ad)i]/(c^2+d^2)
Consider (AC + BD) as a vector multiplied by B vector, and C ^ 2 + D ^ 2 as the module of B vector
(AC + BD) / (C ^ 2 + D ^ 2) = 1 is a * B / (b module)
But I don't know about model a either.... (BC AD) / (C ^ 2 + D ^ 2) = √ 3 will not be used..... But it should be combined with a * B / (a module * B module) = cos ∠ AOB of plane vector
Put it away
The relationship between the modulus and the argument of the quotient of the complex number and the divisor, the modulus and the argument of the divisor
The modulus of the complex quotient is equal to the modulus of the divisor divided by the modulus of the divisor
The argument of the complex quotient is equal to the argument of the divisor minus the argument of the divisor plus 2npi
What is modulus of complex number?
Let z = a + bi (a, B ∈ R)
Then the module of complex Z | Z | = √ A & sup2; + B & sup2;,
Its geometric meaning is the distance from a point (a, b) on the complex plane to the origin
I wish you a happy study!
For example, the complex number: a + IB, √ (a ^ 2 + B ^ 2) is the module of the complex number.
Similar to the module of vector in coordinate form, it is square a + square B under the root sign. A is the coefficient of real part and B is the coefficient of imaginary part.
Square root (the square of the real part plus the square of the imaginary part)
In geometric sense, the distance from the imaginary number to the origin
How to find the modulus of complex number
How to find the module of complex number 5 + 6I
A + bi, a and B are real numbers
Then the module | a + bi | = √ (A & sup2; + B & sup2;)
So | 5 + 6I | = √ (5 & sup2; + 6 & sup2;) = √ 61
Let | Z + 4-3i | - 2 | = 2 - | Z + 4-3i |, then what is the maximum and minimum of | Z |?
The maximum value is 7, and the minimum value is 3. This should be done in a geometric sense,