1. (22+5i) + (3+19i)
2. (4+7i) - (5+i)

Question

1. (22+5i) + (3+19i) 
2. (4+7i) - (5+i)

anonymous · Answer

Just add/ substract constant with constant and i terms together:
(22+5i) + (3+19i) = 25 + 24i 

(4+7i) - (5+i) = -1+ 6i

anonymous · Answer

wat about this -> state the values of the a and b components of a + bi form for each complex. state whether the value is the real partor the imaginary part of the number. 
1. 3-2i 
2. -2-i
3. 3i
4. 0+ 0i
5. -7-4i   plzz

anonymous · Answer

1. a = 2 real , b= -2 imaginary
2. a = -2 real, b = -1 imaginary
similary for the rest

anonymous · Answer

can u show work please how u did it i want to understand

anonymous · Answer

sorry nvm i understand

anonymous · Answer

the form is a + bi
match with 3 - 2i
Can you compare a = ?, b = ?

anonymous · Answer

wat about 2/3 squarerrot of -162

anonymous · Answer

(2/3) sqrt ( -162) 
= (2/3) i *sqrt ( 2 * 9^2)
= (2/3) 9i sqrt(2)

anonymous · Answer

= 6i sqrt (2)

anonymous · Answer

thankss wat about ( 2+5i)^2