-6-4i.
A product is a binary operatoin. That is, it requires two numbers to be combined. There is only one number, 2 + 4i, in the question.
√-48 + 35 + √ 25 + √-27 = √[(i2)(16)(3)] + 35 + 5 + √[(i2)(9)(3)] = 4i√3 + 38 + 9i√3 = 38 + (13√3)i
(7 + 3i) + (8 + 9i) = (7 + 8) + (3i + 9i) = (7 + 8) + (3 + 9)i = 15 + 12i Which can also be written as: 15 + 12i = 3(5 + 4i).
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
-6-4i.
-9
The conjugate of -8-4i is -8+4i. It is obtained by changing the sign of the imaginary part of the complex number.
To get the conjugate simply reverse the sign of the complex part. Thus conj of 7-4i is 7+4i
4i(-2 -3i) = 4i×-2 - 4i×-3i = -8i -12i² = -8i + 12 = 12 -8i → the conjugate is 12 + 8i
(2 + 4i) - (7 + 4i) = -5 2 + 4i - 7 + 4i = -5 + 8i
('|x|' = Absolute value of x) |3+4i| = √(32 + 42) = √(9+16) = √25 = 5 Thus |3+4i| = 5.
Since the imaginary parts cancel, and the real parts are the same, the sum is twice the real part of any of the numbers. For example, (5 + 4i) + (5 - 4i) = 5 + 5 + 4i - 4i = 10.
The multiplicative inverse of a number a is a number b such that axb=1 If we look at (3-4i)/(5+2i), we see that we can multiply that by its reciprocal and the product is one. So (5+2i)/(3-4i) is the multiplicative inverse of (3-4i)/(5+2i)
(3+2i)/(5+4i)If you multiply both sides by the conjugate of the denominator (5-4i), you get:(3+2i)(5-4i)/(5+4i)(5-4i)= (23-2i)/(25 + 16 +20i - 20i)= (23-2i)/41The denominator is now real, because the i terms cancelAs a general formula (easy to expand) this would be:(a+bi)/(c+di) = [(ac+bd) + (bc-ad)i] / (c^2 + d^2)It's a very easy method, but if you're the sort of person who loves using general formulas, there it is.
3-2j.
the problem: what is 4 + 4i + 4 + 6i what you do is add the real and imaginary parts, thus: 4+4 and 4i+6i = 8+10i answer.