Z¹=2-2i z²=6-5i
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14 января 2025 05:59
To analyze the complex numbers ( z^1 = 2 - 2i ) and ( z^2 = 6 - 5i ), you might be interested in various operations such as addition, subtraction, multiplication, division, or finding the modulus and argument of each.
Addition: [ z^1 + z^2 = (2 - 2i) + (6 - 5i) = (2 + 6) + (-2 - 5)i = 8 - 7i ]
Subtraction: [ z^1 - z^2 = (2 - 2i) - (6 - 5i) = (2 - 6) + (-2 + 5)i = -4 + 3i ]
Multiplication: [ z^1 \cdot z^2 = (2 - 2i)(6 - 5i) ] Using the distributive property: [ = 2 \cdot 6 + 2 \cdot (-5i) - 2i \cdot 6 - 2i \cdot (-5i) ] [ = 12 - 10i - 12i + 10i^2 ] Recall that ( i^2 = -1 ): [ = 12 - 10i - 12i - 10 = 2 - 22i ]
Division: [ \frac{z^1}{z^2} = \frac{2 - 2i}{6 - 5i} ] To divide, multiply the numerator and denominator by the conjugate of the denominator: [ = \frac{(2 - 2i)(6 + 5i)}{(6 - 5i)(6 + 5i)} ] Denominator: [ = 6^2 + 5^2 = 36 + 25 = 61 ] Numerator: [ = (2 \cdot 6 + 2 \cdot 5i - 2i \cdot 6 - 2i \cdot 5i) ] [ = 12 + 10i - 12i + 10 = 22 - 2i ] So, [ \frac{z^1}{z^2} = \frac{22 - 2i}{61} = \frac{22}{61} - \frac{2}{61}i ]
Modulus: [ |z^1| = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} ] [ |z^2| = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} ]
Argument: [ \text{arg}(z^1) = \tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1) ] This is in the fourth quadrant, where the angle is ( -\frac{\pi}{4} ) (or ( \frac{7\pi}{4} ) in standard position). [ \text{arg}(z^2) = \tan^{-1}\left(\frac{-5}{6}\right) ] The angle is in the fourth quadrant.
If you have a specific operation or a different aspect of the complex numbers you would like to explore, please let me know!
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