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Math  /  Algebra

Question57. (2+−1)(3−−3)(2+\sqrt{-1})(3-\sqrt{-3})

Studdy Solution

STEP 1

1. We have complex numbers 2+−12 + \sqrt{-1} and 3−−33 - \sqrt{-3}.
2. −1\sqrt{-1} can be replaced by ii, the imaginary unit where i2=−1i^2 = -1.
3. −3\sqrt{-3} can be rewritten as 3⋅−1=3i\sqrt{3} \cdot \sqrt{-1} = \sqrt{3}i.
4. We need to multiply the two complex numbers and simplify the result to be in the form a+bia + bi, where aa and bb are real numbers.

STEP 2

1. Rewrite the given expression using ii for −1\sqrt{-1}.
2. Multiply the two complex numbers.
3. Simplify the product using the properties of ii.
4. Combine like terms to get the final result in the form a+bia + bi.

STEP 3

Rewrite the given expression 2+−12 + \sqrt{-1} and 3−−33 - \sqrt{-3} using ii.
(2+i)(3−3i) (2 + i)(3 - \sqrt{3}i)

STEP 4

Use the distributive property (FOIL method) to multiply the two complex numbers:
(2+i)(3−3i)=2⋅3+2⋅(−3i)+i⋅3+i⋅(−3i) (2 + i)(3 - \sqrt{3}i) = 2 \cdot 3 + 2 \cdot (-\sqrt{3}i) + i \cdot 3 + i \cdot (-\sqrt{3}i)

STEP 5

Multiply each term:
=6−23i+3i−3i2 = 6 - 2\sqrt{3}i + 3i - \sqrt{3}i^2

STEP 6

Simplify the product using the property i2=−1i^2 = -1:
−3i2=−3(−1)=3 -\sqrt{3}i^2 = -\sqrt{3}(-1) = \sqrt{3}
So, we have:
6−23i+3i+3 6 - 2\sqrt{3}i + 3i + \sqrt{3}

STEP 7

Combine the real parts and the imaginary parts:
(6+3)+(−23i+3i) (6 + \sqrt{3}) + (-2\sqrt{3}i + 3i)
=(6+3)+i(3−23) = (6 + \sqrt{3}) + i(3 - 2\sqrt{3})
Solution in the form a+bia + bi is:
6+3+i(3−23) 6 + \sqrt{3} + i(3 - 2\sqrt{3})

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