How do you find the length and direction of vector #-4 - 3i#? Let z=-4-3i z represents a vector on an Argand diagram The magnitude of the vector is the modulus of z, which is found using the Pythagoras theorem |z|=sqrt ( (-4)^2+ (-3)^2)=5 The direction of the vector the principal argument of z, which is found using trigonometry
Does 9isqrt(3)-3i^2sqrt(6) = 3sqrt(6)+3isqrt(3) ? | Socratic Perhaps the #9# should have been under the radical as a multiplier, or more probably there was a preceding step in which #sqrt (-9)# was rewritten as #9i# instead of #sqrt (9)i = 3i# Answer link
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
Find the sum #sum_ (i=1)^6 (3i^2+4i+2)#? - Socratic sum_ (i=1)^6 (3i^2+4i+2)=369 As sum_ (i=1)^n1=n, sum_ (i=1)^ni= (n (n+1)) 2 and sum_ (i=1)^ni^2= (n (n+1) (2n+1)) 6 sum_ (i=1)^n (3i^2+4i+2) =3sum_ (i=1)^ni^2+4sum
How do you simplify # (2 + i) (3 - 5i)#? - Socratic ii-7i These can be multiplied in a manner similar to expanding brackets: (2+i) (3-5i)=6+3i-10i-5i^2 Remember that i = sqrt (-1) so i^2=-1 This will in turn allow us to gather the like terms like so: 6+5+3i-10i=11-7i
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
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