I assume that "I" is a variable2+5i+6+3i7i+6+3i10i+616i is the answer
127
p = (2 + 5i)q = (6 - i)pq = (2 + 5i) (6 - i)pq = 12 + 30i - 2i - 5i2pq = 12 + 28i + 5pq = 17 + 28iIts the same as multiplying polynomials. Just multiply all the combinations of terms, group, and simplify.
Not in real numbers. It can be factored to (x - 5i)(x + 5i) where i is the square root of negative one.
is eassy
1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34
9-5i
1/(2 + 5i) (multiply both the numerator and the denominator by 2 - 5i)= 1(2 - 5i)/(2 + 5i)(2 - 5i)= (2 - 5i)/(4 - 25i2) (substitute -1 for i2)= (2 - 5i)/(4 + 25)= (2 - 5i)/29= 2/29 - (5/29)i
I assume that "I" is a variable2+5i+6+3i7i+6+3i10i+616i is the answer
(x-5i)(x+5i)
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
The conjugate is 7-5i
127
p = (2 + 5i)q = (6 - i)pq = (2 + 5i) (6 - i)pq = 12 + 30i - 2i - 5i2pq = 12 + 28i + 5pq = 17 + 28iIts the same as multiplying polynomials. Just multiply all the combinations of terms, group, and simplify.
Not in real numbers. It can be factored to (x - 5i)(x + 5i) where i is the square root of negative one.
9x2 + 25 has no rational factors. Its factorisation in the complex domain is:(3x + 5i)*(3x - 5i) where i is the imaginary square root of -1.
0 + 5i Its complex conjugate is 0 - 5i