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What is true about the complex number -5-5i?

That 5 is a factor.


Complex conjugate of 5i?

0 + 5i Its complex conjugate is 0 - 5i


What is the usefulness of the conjugate and its effect on other complex numbers?

The conjugate of a complex number is the same number (but the imaginary part has opposite sign). e.g.: A=[5i - 2] --> A*=[-5i - 2] Graphically, as you change the sign, you also change the direction of that vector. The conjugate it's used to solve operations with complex numbers. When a complex number is multiplied by its conjugate, the product is a real number. e.g.: 5/(2-i) --> then you multiply and divide by the complex conjugate (2+i) and get the following: 5(2+i)/(2-i)(2+i)=(10+5i)/5=2+i


What does 5i mean in math form?

In mathematics, ( 5i ) represents a complex number where ( i ) is the imaginary unit, defined as the square root of -1. Thus, ( 5i ) can be understood as a point on the imaginary axis of the complex plane, with a magnitude of 5 and an angle of ( \frac{\pi}{2} ) radians from the real axis. It indicates that there is no real component, only an imaginary one.


What do you call the number that multiplies the variable i?

In the number: 5i, 5 is called the coefficient of i.


What are 10 complex numbers with the absolute value of 5?

Use the Pythagorean theorem. 5, -5, 5i, and -5i will work, as well as any combination of a real and imaginary number such that (real part) squared + (imaginary part) squared = 25, for example, 4 + 3i, 3 + 4i, 4 - 3i, etc.


How do you solve 2 divided by 5i?

0.4


How do you subtract imaginary numbers?

When adding and subtracting complex numbers, you can treat the "i" as any variable. For example, 5i + 3i = 8i, 5i -3i = 2i, etc.; (2 + 5i) - (3 - 3i) = (2 - 3) + (5 + 3)i = -1 + 8i.


What are the square roots of 25?

5, -5, 5i, -5i, 5i^2, and many others... If you're asking this question, chances are it is either 5 or negative 5.


What are the cube roots of 1000i?

5√3 + 5i, -5√3 + 5i, -10i


What is the root of -25?

5i


What is 1 divided by 2 plus 5i in the form a plus bi?

1/(2 + 5i) (multiply both the numerator and the denominator by 2 - 5i)= 1(2 - 5i)/(2 + 5i)(2 - 5i)= (2 - 5i)/(4 - 25i2) (substitute -1 for i2)= (2 - 5i)/(4 + 25)= (2 - 5i)/29= 2/29 - (5/29)i