What is the complex number z=i/(1-i)+i/(1+i)?

neela | Student

z = i/(1-i) +i/(1+i).

To simplify and find the complex number.


The   LCM  of denominators (1-i), (1+i) is (1-i)(1+i) = 1-i^2 = 1+1 =2.

So we know express the give complex number in the  equivalent

form with a common denominator 2, which is the product of (1-i)(1+i).

Therefore z =  i/((1-i)+i/(1+i) = i(1+i)/2 +i(1-i)/2 .

z = (i+i^2)/2 +(i-i^2)/2

z = (i+i^2+i-i^2)/2

z = 2i/2

z = i.

Therefore the given complex number z = i.

giorgiana1976 | Student

To determine the complex number, we'll have to determine the result of the sum of 2 quotients.

To calculate the sum of 2 quotients that do not have a common denominator we'll have to calculate the LCD(least common denominator) of the 2 ratios.

We notice that LCD = (1+i)(1-i)

We notice also that the product (1+i)(1-i) is like:

(a-b)(a+b) = a^2 - b^2

We'll write instead of product the difference of squares, where a = 1 and b = i.

LCD = (1+i)(1-i)

LCD = 1^2 - i^2

We'll write instead of i^2 = -1

LCD = 1 - (-1)

LCD = 2

Now, we'll multiply the first ratio by (1+i) and the second ratio by (1-i):

z = i(1+i)/2 + i(1-i)/ 2

We'll remove the brackets:

z = (i + i^2 + i - i^2)/2

We'll eliminate like terms:

z = 2i/2

We'll simplify:

z =  i

The result is a complex number, whose real part is 0 and imaginary part is 1.

The algebraic form of the complex number z is z = i.

The polar form of the complex number is:

z = |z|(cos a + i*sin a)

We'll calculate the modulus and the argument of z.

|z| = sqrt(Re(z)^2 + Im(z)^2)

|z| = sqrt(0^2 + 1^2)

|z| = 1

tan a = y/x

tan a = 1/0 = +infinite

a = pi/2

The polar form of the complex number is:

z = (cos pi/2 + i*sin pi/2)

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