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
1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).
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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
1/(3+5i)=(3-5i)/((3+5i)(3-5i))=(3-5i)/(9+25)=(3-5i)/34
0.4
4a2+ 25 does not factor over the real number field. In the complex numbers , it factors as (2a +5i)(2a - 5i). This is because i2 = -1, so 4a2 + 25 = 4a2 - (- 25) = 4a2 - 25(-1) = 4a2 - 25i2
0 + 5i Its complex conjugate is 0 - 5i
4
Let a + bi be the reciprocal. So (a + bi)(2 - 5i) = 1 (a + bi)(2 - 5i) = 2a - 5ai + 2bi + 5b = (2a + 5b) + (2b - 5a)i Therefore 2a + 5b = 1 and 2b - 5a = 0. Solving the simultaneous equations, we find that a = 2/29 and b = 5/29. So the reciprocal of 2 - 5i is 2/29 + 5i/29.
Welcome to the world of imaginary and complex numbers. i is defined to be a number such that i2 = -1. i is imaginary ( that is not real) 5i is i+i+i+i+i . There is no simpler form than 5i . Please see the attached link for more about imaginary numbers.
6+5i
12
To find the complex conjugate change the sign of the imaginary part: For 11 + 5i the complex conjugate is 11 - 5i.
You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).You take the additive invers of the real and of the imaginary part. For instance, the additive inverse of: (3 - 5i) is (-3 + 5i).