How to use the plural in English

How to use the plural in English

If it's one, you don't add S. if it's more than one, you need to add S. common singular nouns are: bread, bus, car dog
Some nouns are uncountable and cannot be plural. The corresponding be verb should be used as is, such as water, etc

In the complex plane, it is known that the distance between the complex Z and the point corresponding to the complex number-1-i is 1 In the complex plane, it is known that the distance between the complex Z and the point corresponding to the complex number-1-i is 1. Find Z * conjugate complex Z + (1-I) Z + (1 + I) * conjugate complex Z Detailed process!

Let z = x + iy, then (x + 1) ^ 2 + (y + 1) ^ 2 = 1, and then use the following formula to get x, y, and Z

If the complex Z satisfies | Z | = 1, and the complex number 2Z + 3-4i, then the locus of the corresponding point of the complex number is?

"Lixiedehao" Hello, I'm glad to answer for you!
w=2z+3-4i,
Z = (W-3 + 4I) / 2, because | Z | = 1,
so：|z|=|（w-3+4i）/2|=1,
so：|w-3+4i|=2,
That is, the trajectory of W is a circle: | W-3 + 4I | = 2
I hope my answer will help you~

The design of the complex calculator can do + - * + = - = * = + + - > = C + + complex calculator=

class Complex
{
public:
Complex(){
re=0.0f;
im=0.0f;
I am sorry.
Complex(float _ re,float _ IM)
Note
re = _ Re;
im = _ Im;
I am sorry.
Complex(const Complex &complex)
Note
*this = complex;
I am sorry.
// operators:
Complex &operator=(const Complex &c)
Note
re = c.re;
im = c.im;
return *this;
I am sorry.
Complex operator+(const Complex &c)
Note
return Complex(re+c.re,im+c.im);
I am sorry.
Complex operator-(const Complex &c)
Note
return Complex(re-c.re,im-c.im);
I am sorry.
Complex operator*(const Complex &c)
Note
return Complex(re*c.re - im*c.im,re*c.im + im*c.re);
I am sorry.
Complex operator+=(const Complex &c)
Note
*this = *this+c;
return *this;
I am sorry.
Complex operator-=(const Complex &c)
Note
*this = *this-c;
return *this;
I am sorry.
Complex operator*=(const Complex &c)
Note
*this = *this*c;
return *this;
I am sorry.
 / / + + + +, add 1 to both the virtual and the real
 / / front
Complex &operator++(void)
Note
Wei
return *this = *this+Complex(1,1);
I am sorry.
Complex &operator--(void)
{
return *this = *this-Complex(1,1);
I am sorry.
 / / post
Complex operator++(int t)
Note
t =0;
Complex temp = *this;
++*this;
return temp;
I am sorry.
A kind of
Complex operator--(int t)
Note
t =0;
Complex temp = *this;
--*this;
return temp;
I am sorry.
bool operator==(const Complex &c)
Note
return (re==c.re && im==c.im);
I am sorry.
bool operator!=(const Complex &c)
Note
return !(*this==c);
I am sorry.
 / / module comparison? The following are all
bool operator>(const Complex &c)
Note
return ((re*2+im*2) > (c.re*2+c.im*2));
I am sorry.
bool operator=(const Complex &c)
Note
return !(*thisb)
a=Complex(1,1);
if(a=b)
a=Complex(1,1);
if(a

sin30°cos60°+cos30°sin60°= I can't do it

One

Cos0.cos30.cos60.sin0.sin30.sin60

cos0°=1
cos30°=√3/2
cos60°=1/2
sin0°=0
sin30°=1/2
sin60°=√3/2

Absolute value of root sign (COS 60 ° - 1) 2 + sin 60 ° - Tan 2 emergency

=( √3/2)²-1×1/2-(√3)²
=3/4-1/2-3
=-2 and 3 / 4
Absolute value of root sign (cos60 ° - 1) 2 + sin60 ° - 1
=1-cos60°+1-sin60°
=1-1/2+1-√3/2
=3/2-√3/2

cos²45°-1/cos60°+1/tan45°+cos²30°+sin²45°

The original formula = (√ 2 / 2) ^ 2-1 / (1 / 2) + 1 / 1 + (√ 2 / 2) ^ 2
=(1/2)-2+1+(1/2)
=1-2+1
=0
Hope to adopt, thank you

Sin30 ° + sin60 ° = sin90?

no
sin30°+sin60°=1/2+√3/2=(1+√3)/2
sin90°=1

What is sin 30 degrees plus sin 45 degrees plus sin 90 degrees

3 + root 2 / 2