Question

Find the modulus and argument of the following complex numbers and hence express each of them in the poloar form :

$\left(i\right)1+i\left(ii\right)\sqrt{3}+i\left(iii\right)1-i\left(iv\right)\frac{1-i}{1+i}\left(v\right)\frac{1}{1+i}\left(vi\right)\frac{1+2i}{1-3i}\left(vii\right)sin{120}^0-icos{120}^0\left(viii\right)\frac{-16}{1+i\sqrt{3}}$

Answer 1

(i) 1 + i

The polar form of a complex number z = x + iy, is given by $z=|z|cos\theta +isin\theta$

where,

$|x|=\sqrt{x^2+y^2}andarg\left(z\right)=\theta =ta{n}^{-1}\left(\frac{b}{a}\right)letz=1+i|z|=\sqrt{1^2+1^2}=\sqrt{2}\therefore x,y>0,so\theta \text{lies is first quadrant}\text{Now,}\theta =ta{n}^{-1}\left(\frac{b}{a}\right)=ta{n}^{-1}\left(\frac{1}{1}\right)[\because a=1andb=1]=ta{n}^{-1}\left(1\right)(\because tan\frac{\pi }{4}=1)=\frac{\pi }{4}(\because ta{n}^{-1}(tanx)=x)\Rightarrow arg\left(x\right)=\frac{\pi }{4}$

Polar form: of 1 + i is given by z = $\sqrt{2}$

$(cos\frac{\pi }{4}+isin\frac{\pi }{4})$

$\left(ii\right)\sqrt{3}+i14$

The polar form of a complex numer z = x + iy, is given by z = |z| $(cos\theta +isin\theta )$

where,

$|z|=\sqrt{x^2+y^2}andarg\left(z\right)=\theta =ta{n}^{-1}\left(\frac{b}{a}\right)letz=\sqrt{3}+i|z|=\sqrt{(\sqrt{3}{)}^2+(1{)}^2}=\sqrt{3+1}=\sqrt{4}=2\therefore x=\sqrt{3}>0andy=1>0,\therefore \theta \text{lies in first quadrant}\text{Hence,}\theta =arg\left(z\right)=ta{n}^{-1}\left(\frac{y}{x}\right)=ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)=ta{n}^{-1}\left(\frac{tan\pi }{6}\right)=ta{n}^{-1}(\because ta{n}^{-1}(tanx)=x)\text{Polar form is given by}z=|z|(cos\theta +isin\theta )i.e.z=2(cos\frac{\pi }{6}+isin\frac{\pi }{6})$

$\left(iii\right)1-i\text{Modulus,}|1-i|=\sqrt{1^2+1^2}=\sqrt{2}\text{Argument,}arg(1-i)=ta{n}^{-1}\left(\frac{-1}{1}\right)=ta{n}^{-1}=-\frac{\pi }{4}\text{Polar form,}\sqrt{2}(cos\frac{\pi }{4}-isin\frac{\pi }{4})$

$\left(iv\right)\frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(1+i)(1-i)}=\frac{(1-i{)}^2}{1^2-i^2}=\frac{1-2i-1}{1+1}=\frac{-2i}{2}=-i\text{Modulus,}\left|\frac{1-i}{1+i}\right|=|-i|=1\text{Argument,}ta{n}^{-1}\left(\frac{-1}{0}\right)=-\frac{\pi }{2}\text{Polar Form,}z=r(cos\theta +isin\theta )z=(cos\frac{\pi }{2}-isin\frac{\pi }{2})$

$\left(v\right)\frac{1}{1+i}\text{Modulus,}\left|\frac{1-i}{1+i}\right|=\left|\frac{1(1-i)}{(1+i)(1-i)}\right|\text{[Rationalizing the denominator]}=\left|\frac{1-i}{1^2-i^2}\right|=\left|\frac{1-i}{2}\right|=\sqrt{{\left(\frac{1}{2}\right)}^2+{(-\frac{1}{2})}^2}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}}\text{Argument,}ta{n}^{-1}(-1)=-\frac{\pi }{4}\therefore \text{Polar Form}=\frac{1}{\sqrt{2}}cos\left(\frac{\pi }{4}\right)-isin\left(\frac{\pi }{4}\right)$

$\left(vi\right)\frac{1+2i}{1-3i}\text{The polar form of a complex number}z=x+iy,\text{is given by}z=|z|(cos\theta +isin\theta )\text{where,}|z|=\sqrt{x^2+y^2}andarg\left(z\right)=\theta =ta{n}^{-1}\left(\frac{b}{a}\right)\text{let}z=\frac{1+2i}{1-3i}=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}=\frac{1(1+3i)+2i(1+3i)}{1^2+3^2}=\frac{1+3i+2i-6}{1+9}=\frac{-5+5i}{10}=\frac{-5}{10}+\frac{5}{10}i=\frac{-1}{2}+\frac{1}{2}i\therefore |z|=\sqrt{{\left(\frac{-1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}\text{Here x}=\frac{-1}{2}<0\text{and}y=\frac{1}{2}>0,\therefore \theta \text{lies in quadrant II}\theta =arg\left(z\right)=ta{n}^{-1}\frac{\frac{1}{2}}{\frac{-1}{2}}=ta{n}^{-1}(-1)=ta{n}^{-1}(-tan\frac{\pi }{4})=ta{n}^{-1}\left(tan(\pi -\frac{\pi }{4})\right)[\because tan(\pi -\theta )=-tan\theta ]=\pi -\frac{\pi }{4}=\frac{3\pi }{4}\text{The polar form is given by z}=\frac{1}{\sqrt{2}}(cos\frac{3\pi }{4}+isin\frac{3\pi }{4})$

$\left(vii\right)sin{120}^0-icos{120}^0\text{The polar form of a complex number}z=x+iy,\text{is given by}z=|z|(cos\theta +isin\theta )\text{where,}|z|=\sqrt{x^2+y^2}\text{and}arg\left(z\right)=\theta =ta{n}^{-1}\left(\frac{b}{a}\right)\text{let}z=sin{120}^0-icos{120}^{\circ }=sin(\frac{\pi }{2}+\frac{\pi }{6})-icos(\frac{\pi }{2}+\frac{\pi }{6})(\therefore {120}^{\circ }=\frac{\pi }{2}+\frac{\pi }{6})\Rightarrow z=cos\frac{\pi }{6}+isin\frac{\pi }{6}(\because sin(\frac{\pi }{2}+\theta )=cos\theta andcos(\frac{\pi }{2}+\theta )=-sin\theta )$

Here z is already inpolar form

$with|z|=1and\theta =arg\left(z\right)=\frac{\pi }{6}$

The polar form is given by z

$=(cos\frac{\pi }{6}+isin\frac{\pi }{6})$

$\left(viii\right)\frac{-16}{1+i\sqrt{3}}\text{The polar form of complex numer z = x + iy,is given by}z=|z|(cos\theta +isin\theta )\text{where,}|z|=\sqrt{x^2+y^2}\text{and}arg\left(z\right)=\theta =ta{n}^{-1}\left(\frac{b}{a}\right)letz=\frac{-16}{1+i\sqrt{3}}=\frac{-16}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}=\frac{-16(1-i\sqrt{3})}{(1{)}^2+(\sqrt{3}{)}^2}=\frac{-16(1-i\sqrt{3})}{1+3}=\frac{-16}{4}(1-i\sqrt{3})=-4(1-i\sqrt{3})=-4+4\sqrt{3}i\therefore |z|=\sqrt{(-4{)}^2+(4\sqrt{3}{)}^2}=\sqrt{16+48}=sqrt64=8\text{Here}x=-4<0andy=4R3>0,\therefore \theta \text{lies in quadrant II}\theta =arg\left(z\right)=ta{n}^{-1}\left(\frac{4\sqrt{3}}{-4}\right)=ta{n}^{-1}(-\sqrt{3})={\tan }^{-1}(-tan\frac{\pi }{3})(\because tan(\pi -\theta )=-tan\theta )=\pi -\frac{\pi }{3}=\frac{2\pi }{3}\text{The polar form is given by z = 8}(cos\frac{2\pi }{3}+isin\frac{2\pi }{3})$