The conjugate of the complex number $ \frac {(1+i)^2}{1-i}$ is
- $1 - i$
- $1 + i$
- $-1+ i$
- $-1 - i$
The Correct Option is D
Solution and Explanation
$= \frac{\left(1+ i^{2} +2 i\right)}{1-i} \times \frac{1+i}{1+i} $
$ = \frac{2i\left(1+i\right)}{1-i^{2}} $
$ = \frac{2 i +2i^{2}}{1+1} =\frac{2i-2}{2} $
= i - 1
$\therefore$ Required conjugate is $ -1-i$
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Concepts Used:
Conjugate of a Complex Number
A complex conjugate of a complex number is equivalent to the complex number whose real part is identical to the original complex number and the magnitude of the imaginary part is identical to the opposite sign.
A complex number is of the expression a + ib,
where,
a, b = real numbers, ‘a’ is named as the real part, ‘b’ is named as the imaginary part, and ‘i’ is an imaginary number equivalent to the root of negative 1.
The complex conjugate of a + ib with real part 'a' and imaginary part 'b' is stated by a - ib whose real part is 'a' and imaginary part is '-b'.
a - ib is the reflection of a + ib with reference to the real axis (X-axis) in the argand plane.