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Home » Find the modulus and argument of the following complex numbers and hence express each of them in the polar form: (i) 1/(1 + i) (ii) (1 + 2i) / (1 – 3i)
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1/(1 + i)
(ii) (1 + 2i) / (1 – 3i)
CBSE | Class 11 | Complex Numbers | Exercise 13.4 | Maths | RD Sharma
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
(i) 1/(1 + i)
(ii) (1 + 2i) / (1 – 3i)

Solution:

It is known to us that the polar form of a complex number Z=x+i iy is given by Z=|Z|(\cos \theta+i \sin \theta)
In which,
\begin{array}{l} \left.|Z|=\text { modulus of complex number }=\sqrt{(} x^{2}+y^{2}\right) \\ \theta=\arg (z)=\text { argument of complex number }=\tan ^{-1}(|y| /|x|) \end{array}

(i) 1 /(1+\mathrm{i})
Given that Z=1 /(1+i)
Let’s multiply and divide by (1-\mathrm{i}), we get
\begin{aligned} Z &=\frac{1}{1+i} \times \frac{1-i}{1-i} \\ =\frac{1-i}{1^{2}-i^{2}} \\ =\frac{1-i}{1-(-1)} \\ =\frac{1-i}{2} \end{aligned}
Therefore now,
|Z|=\sqrt{\left(x^{2}+y^{2}\right)}
=\sqrt{\left((1 / 2)^{2}+(-1 / 2)^{2}\right)}
=\sqrt{(1 / 4+1 / 4)}
=\sqrt{(2 / 4)}
=1 / \sqrt{2}
\theta=\tan ^{-1}(|y| /|x|)
=\tan ^{-1}((1 / 2) /(1 / 2))
=\tan ^{-1} 1
As x>0, y<0 complex number lies in 4^{\text {th }} quadrant and the value of \theta is -90^{0} \leq \theta \leq 0^{\circ}.
\begin{array}{l} \theta=-\pi / 4 \\ Z=1 / \sqrt{2}(\cos (-\pi / 4)+i \sin (-\pi / 4)) \\ =1 / \sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4)) \end{array}
As a result, polar form of 1 /(1+i) is 1 / \sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4))

(ii) (1+2 i) /(1-3 i)
Given that Z=(1+2 i) /(1-3 i)
Let’s multiply and divide by (1+3 i), we get
\begin{aligned} Z =\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i} \\ =\frac{1(1+3 i)+2 i(1+3 i)}{1^{2}-(3 i)^{2}} \\ =\frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}} \\ =\frac{1+5 i+6(-1)}{1-9(-1)} \\ =\frac{-5+5 i}{10} \\ =\frac{-1+i}{2} \end{aligned}
Therefore now,
\begin{array}{l} |\mathrm{Z}|=\sqrt{\left(x^{2}+y^{2}\right)} \\ \left.=\sqrt{(}(-1 / 2)^{2}+(1 / 2)^{2}\right) \\ =\sqrt{(1 / 4+1 / 4)} \\ =\sqrt{(2 / 4)} \\ {=} &{1 / \sqrt{2}} \\ {\theta} &{=\tan ^{-1}(|\mathrm{y}| /|\mathrm{x}|)} \\ {=\tan ^{-1}((1 / 2) /(1 / 2))} \\ {=\tan ^{-1} 1} \\ = \end{array}
As x<0, y>0 complex number lies in 2^{\text {nd }} quadrant and the value of \theta is 90^{0} \leq \theta \leq 180^{\circ}
\begin{array}{l} \theta=3 \pi / 4 \\ Z=1 / \sqrt{2}(\cos (3 \pi / 4)+i \sin (3 \pi / 4)) \end{array}
As a result, polar form of (1+2 i) /(1-3 i) is 1 / \sqrt{2}(\cos (3 \pi / 4)+i \sin (3 \pi / 4))

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