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)#.

x = +-5/3i Given: 11x^2+10 = 2x^2-15 Subtract the right hand side from the left to get: 9x^2+25 = 0 Now x^2 >= 0 for all real values of x, so this has no real solutions.

Explanation: Given a complex number #a+bi#, that number's complex conjugate, denoted #bar (a+bi)#, is given by

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?

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

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

The conjugate is =1-3i Let z=a+ib be a complex number The conjugate is barz=a-ib Therefore, z*barz= (a+ib) (a-ib) =a^2-i^2b^2 But i^2=-1 therefore z*barz=a^2+b^2