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_MATH1013_1ABCD_2025 Subsection 2.2 (closed on 20 Sep)

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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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The problem provides the modulus d = 13 of the complex number z and the imaginary part Im(z) = -5. Let z = x + iy. Given Im(z) = -......Login to view full explanation

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( 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 ) = ⋯

Find the standard form of the complex number with modulus [math]2 and argument [math]\dfrac {\pi }{5}. (Correct the answer to 2 decimal places.)

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