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
It is called a complex number.
Yes, a+bi is standard form for a complex number. The numbers (a) and (b) are both real and i is √(-1)
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
"a + bi" is a common way to write a complex number. Here, "a" and "b" are real numbers.Another common way to write a complex number is in polar coordinates - basically specifying the distance from zero, and an angle.
The concept of conjugate is usually used in complex numbers. If your complex number is a + bi, then its conjugate is a - bi.
The standard form of a complex number is a+bi. So the standard form of the negative square root of 5i is 0-√(5i).
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.
7
A number of the form (a + bi) is a complex number.
complex
It is called a complex number.
Yes, a+bi is standard form for a complex number. The numbers (a) and (b) are both real and i is √(-1)
a-bi a(bi)-1 not negative bi
a complex number
It is 3/13 - 2/13*i
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
plus a, bi, cj