No, it is an imaginary number.
"Conjugate" usually means that in one of two parts, the sign is changed - as in a complex conjugate. If the second part is missing, the conjugate is the same as the original number - in this case, 100.
The conjugate of 2 + 3i is 2 - 3i, and the conjugate of 2 - 5i is 2 + 5i.
The complex conjugate of 2-3i is 2+3i.
0 + 5i Its complex conjugate is 0 - 5i
8 - 8i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
88
8i and -8i both satisfy this: (8i)² = (8²)(i²) = (64)(-1) = -64, and (-8i)² = (-8²)(i²) = (64)(-1) = -64
It is just 11+8i.
The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....The i in Oracle 8i and Oracle 9i stands for INTERNET....
To divide by a complex number, write it as a fraction and then multiply the numerator and denominator by the complex conjugate of the denominator - this is formed by changing the sign of the imaginary bit of the number; when a complex number (a + bi) is multiplied by its complex conjugate the result is the real number a² + b² which can be divided into the complex number of the numerator: (-4 - 3i) ÷ (4 + i) = (-4 - 3i)/(4 + i) = ( (-4 - 3i)×(4 - i) ) / ( (4 + i)×(4 - i) ) = (-16 + 4i - 12i + 3i²) / (4² + 1²) = (-16 - 8i - 3) / (16 + 1) = (-19 - 8i)/17
it is having more advantages than 8i
this is a very good question. lets solve (2+3i)/(4-2i). we want to make 4-2i real by multiplying it by the conjugate, or 4+2i (4-2i)(4+2i)=16-8i+8i+4=20, now we have (2+3i)/20 0r 1/10 + 3i/20 notice that -2i times 2i = -4i^2 =-4 times -1 = 4
-5-7=-126i-8i=-2i-12-2i is your final answer
It is celep9i.,-['-['-[' 8i
No, It is not.