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Question #b8278 - Socratic Perhaps if we rewrite this as: (3+2i)* (1-3i) then we mulitply 3 from the first paranthesis with the both constituents from the second We do the same thing for the second constituent from the first paranthesis
Question #bea27 - Socratic (a) Graph is available in the explanation (b) Modulus is ~~5 8, after rounding to the tenths place Given: the Complex Number : (5-3i) A complex number is of the form a+bi So, for our complex number ** (5-3i), a=5 and b = (-3) a Graph of the Complex Number is below: b Modulus of the Complex Number is: A Complex number of the form a+bi, The Modulus of a Complex number can be found using
How do you combine like terms in #6- ( 4- 3i ) - ( - 2- 10i )#? See the entire solution process below: First, remove all of the terms from parenthesis Be careful to handle the signs of each individual term correctly: 6 - 4 + 3i + 2 + 10i Next, group like terms: 10i + 3i + 6 - 4 + 2 Now, combine like terms: (10 + 3)i + (6 - 4 + 2) 13i + 4
Question #dabf9 - Socratic The real part is =5 13 We need (a-b) (a+b)=a^2-b^2 i^2=-1 The conjugate of (a+ib) is (a-ib) We multiply numerator and denominator by the conjugate of the denominator ( (4-i) (2-3i)) ( (2+3i) (2-3i))= (8-12i-2i+3i^2) (4-9i^2) = (5-14i) (13) =5 13-14 13i The real part is =5 13