Anthonyk2013
Junior Member
- Joined
- Sep 15, 2013
- Messages
- 132
Z1-Z2 Z1=2+j4 Z2=3-j
(2-3)+(j4-j)
=-1+j3
is the above correct?
(2-3)+(j4-j)
=-1+j3
is the above correct?
NO!Z1-Z2
Z1=2+j4 & Z2=3-j
(2-3)+(j4-j)
=-1+j3
is the above correct?
NO!
\(\displaystyle (2+4i)-(3-i)=(2-3)+(4 +1 )i=~?\)
How does -i change to +i?
Anthony, use "i" for \(\displaystyle \sqrt{-1}.\)Don't type "j4," "j3," etc. \(\displaystyle \ \ \) Those are not products of a variable and a constant. Instead, type the constant in front as it should be the coefficient for the variable.It's better to put your equations on separate lines, especially if you are unable to place more spaces between them.I understand your limitations in typing underscores in regular typing, as you typed"Z1" instead of "\(\displaystyle z_1,\)" for instance.\(\displaystyle z_1 - z_2\)\(\displaystyle z_1 = 2 + 4i\)\(\displaystyle z_2 = 3 - i\)\(\displaystyle (2 - 3) - (4 - (-1))i\)\(\displaystyle \cdot\)\(\displaystyle \cdot\)\(\displaystyle \cdot\)Z1-Z2 Z1=2+j4 Z2=3-j (2-3)+(j4-j)=-1+j3is the above correct?