0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
How do you solve 2 divided by 5i?
0.4
What are the cube roots of 1000i?
5√3 + 5i, -5√3 + 5i, -10i
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 is the complex conjugate of a plus bi?
The complex conjugate of a+bi is a-bi. This is written as z* where z is a complex number. ex. z = a+bi z* = a-bi r = 3+12i r* = 3-12i s = 5-6i s* = 5+6i t = -3+7i = 7i-3 t* = -3-7i = -(3+7i)
What is CCXXVI in Roman Numerals?
CCXXVI is 226C = 100 & there are 2 of them.X = 10 & there are 2 of them also.V = 5I = 1200 + 20 + 5 + 1 = 226
Write the complex number 2 - i divided by 5 plus i in the usual a plus bi form?
7
How do you solve 2 divided by 5i?
0.4
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
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 is the complex conjugate of -5 plus 7i?
-5 - 7i
What is the absolute value of 7i?
The absolute value is sqrt(72 + 12) = sqrt(49 + 1) = sqrt(50) or 5*sqrt(2) = 7.071 approx.
Complex conjugate of 5i?
0 + 5i Its complex conjugate is 0 - 5i
What is the reciprocal of 2-5i?
Let a + bi be the reciprocal. So (a + bi)(2 - 5i) = 1 (a + bi)(2 - 5i) = 2a - 5ai + 2bi + 5b = (2a + 5b) + (2b - 5a)i Therefore 2a + 5b = 1 and 2b - 5a = 0. Solving the simultaneous equations, we find that a = 2/29 and b = 5/29. So the reciprocal of 2 - 5i is 2/29 + 5i/29.
What are the cube roots of 1000i?
5√3 + 5i, -5√3 + 5i, -10i
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 is the root of -25?
5i
Express this complex number in standard form i x sum of 5 plus 2i x sum of i-1?
i(2i+5)(i-1) [I hope I've interpreted this correctly - WikiAnswers does not make it easy to pose mathematical questions!] First of all, multiply out in ordinary fashion: i(2i+5)(i-1) = i(2i2 + 3i - 5) = 2i3 + 3i2 - 5i Now, i is defined in this way: i2 = -1 Multiplying both sides by i we get: i3 = -i Substitute this information into the expression: 2i3 + 3i2 - 5i = -2i - 3 - 5i = -3 - 7i Done. ---- As the exponent of i is raised, incidentally, you will observe a cycle like this: i, -1, -i, 1, i, -1...
