Question:
The modulus and amplitude of $\frac{ 1 + 2i}{1 - (1 - i)^2}$ are respectively
The modulus and amplitude of $\frac{ 1 + 2i}{1 - (1 - i)^2}$ are respectively
Updated On: May 12, 2024
- $ 1, \frac{\pi}{3}$
- $ \sqrt{2} , \frac{\pi}{6}$
- $1, 0$
- $\sqrt{3} , 0$
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The Correct Option is C
Solution and Explanation
Let $Z = \frac{ 1 + 2i}{1 - (1 - i)^2} $
$= \frac{1+2i}{1-\left(1 + i^{2} - 2i\right)} = \frac{1+2i}{1+2i} = 1 \Rightarrow Z = 1 + 0i $
Now, put $1 = r \cos \theta , 0 = r \sin \theta$
$ r= \sqrt{1+0} = 1 \Rightarrow \left|Z\right| = \sqrt{a^{2} + b^{2}} = \sqrt{1^{2}+0} = 1 $
$\therefore \:\:\: \tan \theta = 0 \Rightarrow \theta = 0 $
$ \therefore$ Modulus is 1 and amplitude is 0.
$= \frac{1+2i}{1-\left(1 + i^{2} - 2i\right)} = \frac{1+2i}{1+2i} = 1 \Rightarrow Z = 1 + 0i $
Now, put $1 = r \cos \theta , 0 = r \sin \theta$
$ r= \sqrt{1+0} = 1 \Rightarrow \left|Z\right| = \sqrt{a^{2} + b^{2}} = \sqrt{1^{2}+0} = 1 $
$\therefore \:\:\: \tan \theta = 0 \Rightarrow \theta = 0 $
$ \therefore$ Modulus is 1 and amplitude is 0.
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.