Complex Numbers and Quadratic Equations Test 49

Let,

\(z=\frac{1+i}{1-i}\)

Multiplying by \((1+ i)\) in numerator and denominator, we get

\(z=\frac{1+ i }{1- i } \times \frac{1+ i }{1+ i } \)

\(\Rightarrow z=\frac{(1+ i )^{2}}{1^{2}- i ^{2}} \)

\(\Rightarrow z=\frac{1+ i ^{2}+2i}{1-i^2}\)

\(\Rightarrow z=\frac{1+ (-1)+2i}{1-(-1)} \quad[\because i^2=-1] \)

\(\Rightarrow z=\frac{0-2i}{1+1} \)

\(\Rightarrow z=\frac{-2 i }{2}\)

\(\Rightarrow z= -i\)

So, Conjugate of \(z=\bar{z}= i\)

As we know,

If \(z=x+i y\)...(1) be any complex number, then its argument is given by,

\(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{y}{x}\right)\)

Let,

\(z=\frac{1+ i }{1- i }\)

Multiplying by \((1+i)\) in numerator and denominator, we get

\(z=\frac{1+ i }{1- i } \times \frac{1+ i }{1+ i } \)

\(\Rightarrow z=\frac{(1+ i )^{2}}{1^{2}- i ^{2}} \)

\(\Rightarrow z=\frac{1+ i ^{2}+2i}{1-i^2}\)

\(\Rightarrow z=\frac{1+ (-1)+2i}{1-(-1)} \quad[\because i^2=-1] \)

\(\Rightarrow z=\frac{0-2i}{1+1} \)

\(\Rightarrow z=\frac{2 i }{2}\)

\(\Rightarrow z= i\)...(2)

Comparing (1) and (2), we get

\(x+i y=i\)

So, \(x=0\) and \(y=1\)

As, \(\operatorname{Arg}(z)=\tan ^{-1}\left(\frac{y}{x}\right)\)

\(\therefore \operatorname{Arg}(z)=\tan ^{-1}\left(\frac{1}{0}\right)\)

\(\Rightarrow \operatorname{Arg}(z)=\tan ^{-1}(\infty)\)

\(\Rightarrow \operatorname{Arg}(z)=\frac{\pi}{2}\)

Given,

\(z _{1}=6+2 i\)...(1)

\(z _{2}=2- i\)...(2)

On dividing equation (1) and (2), we get

\(\frac{ z _{1}}{ z _{2}}=\frac{6+2 i }{2- i }\)

Multiplying by \((2+i)\) in numerator and denominator, we get

\(\frac{ z _{1}}{ z _{2}}=\frac{6+2 i }{2- i } \times \frac{2+ i }{2+ i } \)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=\frac{12+6 i +4 i +2 i ^{2}}{2^{2}- i ^{2}} \)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=\frac{12+10 i +2\times (-1)}{4-(-1)}\quad[\because i^2=-1] \)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=\frac{12+10 i -2}{4+1}\)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=\frac{10+10 i }{5} \)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=\frac{10(1+ i )}{5} \)

\(\Rightarrow \frac{ z _{1}}{ z _{2}}=2(1+ i )\)

Given,

\(\tan \alpha\) and \(\tan \beta\) are the roots of the equation \(x^{2}+b x+c=0\).

So,

Sum of roots \(= \tan \alpha+\tan \beta=-(-4)=4 \)...(1)

Product of roots \(= \tan \alpha \cdot \tan \beta=-3\)...(2)

As we know,

\(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)

\( \therefore \tan (\alpha+\beta)=\frac{4}{1-(-3)}\)

\(\Rightarrow \tan (\alpha+\beta)=\frac{4}{4}\)

\(\Rightarrow \tan (\alpha+\beta)=1 \)

\(\Rightarrow \tan (\alpha+\beta)=\tan 45^{\circ} \)

\(\therefore \alpha+\beta=45^{\circ}\)

Equality of complex numbers:

Two complex numbers \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=x_{2}+i y_{2}\) are equal if and only if \(x_{1}=x_{2}\) and \(y_{1}=y_{2}\).

or \(\operatorname{Re}\left(z_{1}\right)=\operatorname{Re}\left(z_{2}\right)\) and \(\operatorname{Im}\left(z_{1}\right)=\operatorname{lm}\left(z_{2}\right)\)

Given: \({x}+{iy}=\frac{3+4 {i}}{2-{i}}\)

\(\Rightarrow {x}+{iy}=\frac{3+4 {i}}{2-{i}} \times \frac{2+{i}}{2+{i}}\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}+4 {i}^{2}}{4-{i}^{2}}\)

As we know \({i}^{2}=-1\)

\(\Rightarrow {x}+{iy}=\frac{6+11 {i}-4}{4+1}\)

\(\Rightarrow {x}+{iy}=\frac{2+11 {i}}{5}=\frac{2}{5}+{i} \frac{11}{5}\)

Comparing real and imaginary parts, we get,

\({x}=\frac{2}{5}\) and \({y}=\frac{11}{5}\)

As we know,

If \(z=x+i y\)...(1) be any complex number, then its modulus is given by,

\(| z |=\sqrt{ x ^{2}+ y ^{2}}\)

Let,

\(z =\frac{1+7 i }{(2- i )^{2}} \)

\(=\frac{1+7 i }{2^{2}+ i ^{2}-4 i } \)

\(=\frac{1+7 i }{4-1-4 i } \)

\(=\frac{1+7 i }{3-4 i } \)

\(=\frac{1+7 i }{3-4 i } \times \frac{3+4 i }{3+4 i } \)

\(=\frac{3+4 i +21 i +28 i ^{2}}{3^{2}-(4 i )^{2}} \)

\(=\frac{3+25 i -28}{9+16} \)

\(=\frac{-25+25 i }{25} \)

\(\therefore z=-1+ i \)...(2)

Comparing equation (1) and (2), we get

\(x=-1\) or \(y=1\)

As, \(| z |=\sqrt{ x ^{2}+ y ^{2}}\)

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

\(\Rightarrow|z|=\sqrt{1+1}\)

\(\Rightarrow |z|=\sqrt{2}\)

So, the modulus of \(\frac{1+7 i }{(2- i )^{2}}\) is \(\sqrt{2}\).

Let,

Complex number, \(z=a+i b\)

Conjugate of complex number \(=\bar{z}=a-i b\)

So, \(z \bar{z}=(a+i b)(a-i b) \)

\(\Rightarrow z \bar{z}=a^{2}-(i b)^{2} \)

\(\Rightarrow z \bar{z}=a^{2}-i^{2}(b)^{2} \)

\(\Rightarrow z \bar{z}=a^{2}-(i b)^{2} \)

\(\Rightarrow z \bar{z}=a^{2}+b^{2}\)...(1) \(\quad\left(\because i^{2}=-1\right)\)

As we know,

Modulus of complex number is given by,

\(|z|=\sqrt{(a^{2}+b^{2})}\)

On squaring both sides, we get

\(|z|^{2}=\left(\sqrt{a^{2}+b^{2}}\right)^{2}\)

\(\therefore |z|^{2}= a ^{2}+ b ^{2} \)...(2)

Comparing equation (1) and (2), we get

\( z\bar{z}=| z |^{2}\)

Now,

\( z ^{-1}=\frac{1}{ z }=\frac{1}{ a + ib } \)

\(\Rightarrow z ^{-1}=\frac{1}{ a + ib } \times \frac{ a - ib }{ a - ib }\)

\(\Rightarrow z ^{-1}=\frac{1}{ a + ib } \times \frac{ a - ib }{ a - ib }\)

\(\Rightarrow z ^{-1}=\frac{ a - ib }{ a ^{2}- (ib )^{2}} \)

\(\Rightarrow z ^{-1}=\frac{ a - ib }{ a ^{2}- i^2b^{2}} \)

\(\Rightarrow z ^{-1}=\frac{ a - ib }{ a ^{2}- (-1)b^{2}} \quad[\because i^2=-1]\)

\(\Rightarrow z ^{-1}=\frac{ a - ib }{ a ^{2}+b^{2}}\)

\(\therefore z ^{-1}= \frac{\bar{z}}{| z |^{2}} \neq \frac{ z }{| z |^{2}}\)

So, only statement 1 is correct.

As we know,

For any complex number \(z = x + iy\) the conjugate of \(z\) is given by,

\(z̅ = x - iy\)

Let,

\(z=\frac{3+2 i}{2-i} \)

Multiplying by \((2+i)\) in numerator and denominator, we get

\(z=\frac{3+2 i}{2-i} \times \frac{2+i}{2+i} \)

\(\Rightarrow z=\frac{6+7 i+2 i^{2}}{2^{2}-(i)^{2}} \)

\(\Rightarrow z=\frac{4+7 i}{5} \quad [\because i^2=-1]\)

\(\Rightarrow z=\frac{4}{5}+\frac{7}{5} i\)

\(\therefore\) Conjugate of \(z=\bar{z}=\frac{4}{5}-\frac{7}{5} i\)

Given,

\((-1+i \sqrt{3})^{48}\)

Consider \(z=(x+i y)=(-1+i \sqrt{3})\)

\(\therefore x =1\) and \(y =\sqrt{3}\)

The polar form of the complex number is given by,

\( z = r (\cos \theta+ i \sin \theta)\)

Where \(r =\sqrt{x^{2}+y^{2}}\) and \(\theta=\tan ^{-1}\left(\frac{ y }{ x }\right)\)

So,

\(r=\sqrt{1^{2}+(\sqrt{3})^{2}}=2\)

\(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{1}\right)=\frac{\pi}{3}\)

\(\Rightarrow z=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \)

\(\Rightarrow z^{48}=\left[2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)\right]^{48} \)

\(\Rightarrow z^{48}=2^{48}\left(\cos \left(48 \times \frac{\pi}{3}\right)+i \sin \left(48 \times \frac{\pi}{3}\right)\right. \)

\(\Rightarrow z^{48}=2^{48}[\cos (16 \pi)+i\sin(16 \pi)] \)

\(\Rightarrow z^{48}=2^{48}\left[1+i(0)\right]\)

\(\Rightarrow z^{48}=2^{48}\)

So, the value of \((-1+i \sqrt{3})^{48}=2^{48}\)

Let \(\alpha\) and \(\beta\) are roots of \(a x^{2}+b x+c=0\), then \(\alpha \times \beta=\frac{c}{a}\).

Given,

\(x^{2}+x+2=0\)

Comparing with \(a x^{2}+b x+c=0\), we get

\(a=1, b=1, c=2\)

Product of roots \(=\alpha \times \beta=\frac{c}{a}=\frac{2}{1}=2\)

Now,

\(\frac{\alpha^{10}+\beta^{10}}{\alpha^{-10}+\beta^{-10}}=\frac{\alpha^{10}+\beta^{10}}{\frac{1}{\alpha^{10}}+\frac{1}{\beta^{10}}} \)

\(=\frac{\alpha^{10}+\beta^{10}}{\frac{\beta^{10}+\alpha^{10}}{\alpha^{10} \cdot \beta^{10}}} \)

\(=\alpha^{10} \cdot \beta^{10} \)

\(=(\alpha \cdot \beta)^{10} \)

\(=(2)^{10} \)

\(=1024\)