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

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#.

Apr 6, 2017 · 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 …

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

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 ...

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 …

It's not possible. Let's take a look at the skeleton equation: Li_3PO_4+CsI -> Li_3I+Cs_3PO_4 There is no way we can get this to follow the law of conservation of matter, because cesium is the only …

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

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 …

Explanation: #"note that "sqrt (-9)=3i# #rArr (-3+3i)/6# #= (-3)/6+ (3i)/6=-1/2+1/2i# Answer link