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
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