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SMAT011 Weekly Quiz 3 |LA003

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( 6 โˆ’ 6 ๐‘– โˆ’ 9 3 โˆ’ 27 ๐‘– ) 256 = _ _ _ _ _ _ _ _ _ _ ย  Hints: Convert the complex numbers to polar form. Ifย ย  ๐‘ง = ๐‘Ÿ ( cos โก ๐œƒ + ๐‘– sin โก ๐œƒ ) then ย  ๐‘ง ๐‘› = ๐‘Ÿ ๐‘› ( cos โก ๐‘› ๐œƒ + ๐‘– sin โก ๐‘› ๐œƒ ) . If ๐‘ง 1 = ๐‘Ÿ 1 ( cos โก ๐œƒ 1 + ๐‘– sin โก ๐œƒ 1 ) ย  and ย  ๐‘ง 2 = ๐‘Ÿ 2 ( cos โก ๐œƒ 2 + ๐‘– sin โก ๐œƒ 2 ) then: ย  ย  ย ย ย  ย  ๐‘ง 1 ๐‘ง 2 = ๐‘Ÿ 1 ๐‘Ÿ 2 [ cos โก ( ๐œƒ 1 + ๐œƒ 2 ) + ๐‘– sin โก ( ๐œƒ 1 + ๐œƒ 2 ) ] ย  and ย  ๐‘ง 1 ๐‘ง 2 = ๐‘Ÿ 1 ๐‘Ÿ 2 [ cos โก ( ๐œƒ 1 โˆ’ ๐œƒ 2 ) + ๐‘– sin โก ( ๐œƒ 1 โˆ’ ๐œƒ 2 ) ] .

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The prompt presents a complex-number exponentiation task using polar form, but the provided data for answer choices is incomplete. I will first restate what is visible and then outline the full solution approach, while also explaining what is missing and how to proceed if the options were present. Re-stated question elements: - We are given a complex number expressed in a fragmented form: (6 โˆ’ 6i โˆ’ 9^3 โˆ’ 27i) and then the operation z^256 on that quantity, i.e., (6 โˆ’ 6i โˆ’ 9^3 โˆ’ 27i)^256, with hints about converting to polar form and using De Moivreโ€™s formula: if z = r(cos ฮธ + i sin ฮธ) then z^n = r^n (cos(nฮธ) + i sin(nฮธ)). The hints also show how to multiply arguments when multiplying complex numbers, and how to raise a single complex number to a power. - The target is to fill in blanks corresponding to the resulting expression after raising to the 256th power, presumably in polar for......Login to view full explanation

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