4 i 3  Simplify (3+2i) (i) (3 + 2i) (4 − 3i) (3 + 2 i) (4 - 3 i) Expand (3+2i)(4− 3i) (3 + 2 i) (4 - 3 i) using the FOIL Method. Tap for more steps Apply the distributive property. 3 (4 − 3 i) + 2 i (4 − 3 i) 3 (4 - 3 i) + 2 i (4 - 3 i) Apply the distributive property.

· Multiply by the complex conjugate of the denominator. 3 − i 4 − i (4 +i 4 +i) = 12 + 3i − 4i − i2 16 + 4i − 4i − i2 = 12− i − i2 16− i2. Remember that i = √−1, so i2 = − 1. 12 −i − (− 1) 16− (− 1) = 13− i 17 = 13 17 − 1 17i. Notice that the answer is written in the form of a complex number a + bi. Answer.

Subproblem 3 Set the factor '(4 + -3i)' equal to zero and attempt to solve: Simplifying 4 + -3i = 0 Solving 4 + -3i = 0 Move all terms containing i to the left, all other terms to the right. Add '-4' to each side of the equation. 4 + -4 + -3i = 0 + -4 Combine like terms: 4 + -4 = 0 0 + -3i = 0 + -4 -3i = 0 + -4 Combine like terms: 0 + -4 =

Imaginary Numbers are not "Imaginary". Imaginary Numbers were once thought to be impossible, and so they were called "Imaginary" (to make fun of them).. But then people researched them more and discovered they were actually useful and important because they filled a gap in mathematics but the "imaginary" name has stuck.. And that is also how the name "Real Numbers" came about (real is not.

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3 thoughts on “4 i 3”

1. Mikadal says:

Evaluate (4+i)/(i) Multiply the numerator and denominator of by the conjugate of to make the denominator real. Multiply. Tap for more steps Combine. Simplify the numerator. Tap for more steps Expand using the FOIL Method. Tap for more steps Apply the distributive property.

2. Arashigore says:

(i) / (1+i) = i-3 / (2-i) = i 6i / (4+3i) = +i Absolute value or modulus The absolute value or modulus is the distance of the image of a complex number from the origin in the plane.

3. Daikazahn says:

Calculate i to the power of 16 and get 1. Calculate i to the power of 1 6 and get 1. Re (1-i-3\times 1+4i^ {5}) R e (1 − i − 3 × 1 + 4 i 5) Multiply 3 and 1 to get 3. Multiply 3 and 1 to get 3. Re (1-i-3+4i^ {5}) R e (1 − i − 3 + 4 i 5) Subtract 3 from 1-i to get i.

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