Dec 12, 2017 · We have 4 zeros, 3 with multiplicity 3 and -3,3i and -3i with multiplicity of 1. y= (x^2-9) (x^2+9) (x-3)^2 = (x^2-3^2) (x^2- (3i)^2) (x-3)^2 = (x+3) (x-3) (x-3i) (x+ ...

b) I multiply the 2 brackets: #4*7-4*3i+7*3i-3*3i^2=28-12i+21i+9=37+9i# Answer link

Aug 14, 2017 · Explanation: The first thing we notice with the two expression here is that the denominators are the same since #a^2+9=9+a^2#.

The unit vector has magnitude 1. Divide #A + B# by its magnitude. This means multiplying by the reciprocal of #sqrt (137)#.

One solution of x3 + (2 − i)x2 + (− 4 − 3i)x + (1 + i) = 0 is x = 1 + i. Find the only positive real solution for x?

How can I determine the minimum degree of the equation as well as write the equation in general form given the roots -2i, 3i and -2±i (sqrt^ (3)), 4?

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 ...

For #z_1 = -3 + 2i# and #z_2=4+3i#, write #z_1/z_2# in polar form. In which quadrant will it lie in an Argand Diagram? Precalculus

-156-248i (19-3i)* (-6-14i)=-114+18i-266i-42=-156-248i

Jan 28, 2016 · How do you divide 3i 1 + i + 2 2 + 3i? Precalculus Complex Numbers in Trigonometric Form Division of Complex Numbers 2 Answers Sihan Tawsik