题目
MATH136-1IP-IO-202430-I-81X M04: Midterm Exam Review Quiz
简答题
Simplify the complex numbers: (2−3i)−(3+2i)
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标准答案
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思路分析
We start with the expression (2 − 3i) − (3 + 2i).
First, distribute the subtraction across the second complex number: 2 − 3i − 3 − 2i......Login to view full explanation登录即可查看完整答案
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类似问题
( 6 − 6 𝑖 − 9 3 − 27 𝑖 ) 256 = _ _ _ _ _ _ _ _ _ _ Hints: Convert the complex numbers to polar form. If 𝑧 = 𝑟 ( cos 𝜃 + 𝑖 sin 𝜃 ) then 𝑧 𝑛 = 𝑟 𝑛 ( cos 𝑛 𝜃 + 𝑖 sin 𝑛 𝜃 ) . If 𝑧 1 = 𝑟 1 ( cos 𝜃 1 + 𝑖 sin 𝜃 1 ) and 𝑧 2 = 𝑟 2 ( cos 𝜃 2 + 𝑖 sin 𝜃 2 ) then: 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 [ cos ( 𝜃 1 + 𝜃 2 ) + 𝑖 sin ( 𝜃 1 + 𝜃 2 ) ] and 𝑧 1 𝑧 2 = 𝑟 1 𝑟 2 [ cos ( 𝜃 1 − 𝜃 2 ) + 𝑖 sin ( 𝜃 1 − 𝜃 2 ) ] .
The polar form of 𝑧 = − 6 7 + 3 2 7 𝑖 is: Hint: 𝑧 = 𝑎 + 𝑏 𝑖 = 𝑟 ( cos 𝜃 + 𝑖 sin 𝜃 ) where 𝑟 = 𝑎 2 + 𝑏 2 and 𝜃 = tan − 1 𝑏 𝑎 . Also, don't forget to plot the complex number on the Argand diagram.
Given four complex numbers 𝑧 1 = 2 + 3 𝑖 , 𝑧 2 = − 7 − 5 𝑖 , 𝑧 3 = − 9 + 7 𝑖 , 𝑧 4 = 2 + 5 𝑖 . Calculate | 𝑧 3 ¯ 𝑧 1 + 𝑧 2 𝑧 4 | . Formulae: If 𝑧 = 𝑎 + 𝑏 𝑖 then 𝑧 ¯ = 𝑎 − 𝑏 𝑖 and | 𝑧 | = 𝑎 2 + 𝑏 2 . 𝑎 + 𝑏 𝑖 𝑐 + 𝑑 𝑖 = ( 𝑎 + 𝑏 𝑖 ) ( 𝑐 − 𝑑 𝑖 ) ( 𝑐 + 𝑑 𝑖 ) ( 𝑐 − 𝑑 𝑖 ) = ⋯
The figure shows the Argand diagram together with the complex number \(z\). If \(d=13\) and the imaginary part of \(z\) is \(-5\), find \(z\).
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