complex number

TanishkClass 11, RD sharma

Exercise 13.4

QUESTION 1

Find the modulus and argument of the following complex number and hence express each of them in the polar form :

(i) 1+i

Sol :

z=1+i

r=|z|

=\sqrt{1+1}

=\sqrt{2}

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\Rightarrow \tan \alpha=\left(\frac{1}{1}\right)

\Rightarrow \quad \alpha=\frac{\pi}{4}

Since point (1,1) lies in the first quadrant , the argument of z is given by \theta=\quad \alpha=\frac{\pi}{4}

Polar form =r(\cos \theta+i \sin \theta)

=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)

 

(ii) \sqrt{3}+i

Sol :

z=\sqrt{3}+i

r=|z|

=\sqrt{3+1}

=\sqrt{4}

=2

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\Rightarrow \tan \alpha=\left(\frac{1}{\sqrt{3}}\right)

\Rightarrow \quad \alpha \quad=\frac{\pi}{6}

Since point (\sqrt{3},1) lies in the quadrant , the argument of z is given by \theta=\quad \alpha=\frac{\pi}{6}

Polar form =r(\cos \theta+i \sin \theta)

=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)

 

(iii) 1-i

Sol :

z=1-i

r=|z|

=\sqrt{1+1}

=\sqrt{2}

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\Rightarrow \tan \alpha=\left|\frac{-1}{1}\right|

\Rightarrow =\frac{\pi}{4}

\Rightarrow \alpha=\frac{\pi}{4}

Since point (1,-1) lies in the quadrant , the argument of z is given by \theta=\quad \alpha=-\frac{\pi}{4}

Polar form =r(\cos \theta+i \sin \theta)

=\sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right\}

=\sqrt{2}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)

 

(iv) \frac{1-i}{1+i}

Sol :

Rationalizing the denominator :

=\frac{1-i}{1+i} \times \frac{1-i}{1-i}

\Rightarrow \frac{1+i^{2}-2 i}{1-i^{2}}

\Rightarrow \frac{-2 i}{2} \left(\because i^{2}=-1\right)

\Rightarrow-i

 

r=|z|

=\sqrt{0+1}

= 1

Since point (0,-1) lies on the negative direction of the imaginary axis , the argument of z is given by \frac{3 \pi}{2}

Polar form ==r(\cos \theta+i \sin \theta)

=\left(c \operatorname{os} \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)

=\left\{\cos \left(2 \pi-\frac{\pi}{2}\right)+i \sin \left(2 \pi-\frac{\pi}{2}\right)\right\}

=\left(\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}\right)

 

(v) \frac{1}{1+i}

Sol :

Rationalizing the denominator

=\frac{1}{1+i} \times \frac{1-i}{1-i}

\Rightarrow \frac{1-i}{1-i^{2}}

\Rightarrow \frac{1-i}{2} \left(\because i^{2}=-1\right)

\Rightarrow \frac{1}{2}-\frac{i}{2}

 

r=|z|

=\sqrt{\frac{1}{4}+\frac{1}{4}}

=\frac{1}{\sqrt{2}}

 

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\Rightarrow \tan \alpha=\left|\frac{\frac{\phantom{-}1}{2}}{\frac{-1}{2}}\right|

= 1

\Rightarrow \alpha=\frac{\pi}{4}

Since point (\frac{1}{2},-\frac{1}{2}) lies in the fourth quadrant , the argument is given by \theta=-\alpha=\frac{-\pi}{4}

Polar form =r(\cos \theta+i \sin \theta)

=\frac{1}{\sqrt{2}}\left\{c \operatorname{os}\left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right\}

=\frac{1}{\sqrt{2}}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)

 

(vi) \frac{1+2 i}{1-3 i}

Sol :

Rationalizing the denominator :

\Rightarrow \frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}

\Rightarrow \frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}}

\Rightarrow \frac{-5+5 i}{10} \left(\because i^{2}=-1\right)

\Rightarrow \frac{-1}{\phantom{-}2}+\frac{i}{2}

 

r=|z|

=\sqrt{\frac{1}{4}+\frac{1}{4}}

=\frac{1}{\sqrt{2}}

 

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\tan \alpha=\left|\frac{\frac{1}{2}}{\frac{-1}{2}}\right|

= 1

\Rightarrow \alpha=\frac{\pi}{4}

Since point \left(\frac{-1}{2}, \frac{1}{2}\right) lies in the second quadrant , the argument is given by

\theta=\pi-\alpha

=\pi-\frac{\pi}{4}

=\frac{3 \pi}{4}

Polar form =r(\cos \theta+i \sin \theta)

=\frac{1}{\sqrt{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)

 

(vii) \sin 120^{\circ}-i \cos 120^{\circ}

Sol :

=\frac{\sqrt{3}}{2}+\frac{i}{2}

 

r=|z|

=\sqrt{\frac{3}{4}+\frac{1}{4}}

= 1

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\tan \alpha=\left|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right|

=\frac{1}{\sqrt{3}}

\Rightarrow \alpha=\frac{\pi}{6}

Since point \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right) lies in the first quadrant , the argument is given by \theta=\alpha=\frac{\pi}{6}

Polar form =r(\cos \theta+i \sin \theta)

=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}

 

(viii) \frac{-16}{1+i \sqrt{3}}

Sol :

Rationalizing the denominator:

\Rightarrow \frac{-16}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}

\Rightarrow \frac{-16+16 \sqrt{3} i}{1-3 i^{2}}

\Rightarrow \frac{-16+16 \sqrt{3} i}{4} \left(\because i^{2}=-1\right)

\Rightarrow-4+4 \sqrt{3} i

 

r=|z|

=\sqrt{16+48}

= 8

 

Let \tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|

\tan \alpha=\left|\frac{4 \sqrt{3}}{-4}\right|

=\sqrt{3}

\Rightarrow \alpha=\frac{\pi}{3}

Since the point (-4,4 \sqrt{3}) lies in the third quadrant , the argument is given by

\theta=\pi-\alpha

=\pi-\frac{\pi}{3}

=\frac{2 \pi}{3}

 

Polar form =r(\cos \theta+i \sin \theta)

=8\left\{\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right\}

 

QUESTION 2

Write (i^25)^3 in polar form 

Sol :

\left(i^{25}\right)^{3}=i^{75}

=i^{4 \times 18+3}

=\left(i^{4}\right)^{18} \cdot i^{3}

=i^{3} 

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Let z=0-i

Then |z|=\sqrt{0^{2}+(-1)^{2}}=1

Let \theta be the argument of z and \alpha be the acute angle given by \tan \alpha=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}

Then ,

tan~\alpha=\dfrac{1}{0}=\infty

\Rightarrow \alpha=\frac{\pi}{2}

Clearly,  z  lies in fourth quadrant. So, arg (z)=-\alpha=-\frac{\pi}{2}

The polar form of z is |z|(\cos \theta+i \sin \theta)

=\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)

Thus , the polar form of \left(7^{25}\right)^{3} is \cos \left(\frac{\pi}{2}\right)-i \sin \left(\frac{\pi}{2}\right)

 

QUESTION 3

Express the following complex in the formr(\cos \theta+i \sin \theta)

(i) 1+i \tan \alpha

Sol :

Let \mathrm{z}=1+i \tan \alpha

 

(ii)

Sol :

 

 

(iii)

Sol :

 

 

(iv)

Sol :

 

 

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