1. Evaluate $\left [ i^{18}+\left ( \frac{1}{i} \right )^{25} \right ]^{3}$
= $\left [ (i^{4})^{4}.i^{2}+\frac{1}{(i^{4})^{6}.i} \right ]^{3}$
= $\left [ (1).(-1)+\frac{1}{(1).i} \right ]^{3}$
= $\left [ -1+\frac{1}{i}*\frac{i}{i} \right ]^{3}$
= $\left [ -1+\frac{i}{i^{2}} \right ]^{3}$
= $\left [ -1+\frac{i}{-1} \right ]^{3}$
= $\left [ -1-i \right ]^{3}$
= $(-1)^{3}(1+i)^{3}$
= $-(1+3i+3i^{2}+i^{3})$
= $-(1+3i-3-i)$
= $-(-2+i2)$
= $2-i2$
2. For any two complex numbers $z_{1}$ and $z_{2}$ prove that Re($z_{1}$ $z_{2}$ ) = Re$z_{1}$ Re$z_{2}$- Im$z_{1}$ Im$z_{2}$
Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$
$z_{1}z_{2}$ $=(x_{1}+iy_{1})(x_{2}+iy_{2})$
= $x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x_{2}+i^{2}y_{1}y_{2}$
= $x_{1}x_{2}+i(x_{1}y_{2}+y_{1}x_{2})-y_{1}y_{2}$
= $(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})$
Re($z_{1}$ $z_{2}$ )$=x_{1}x_{2}-y_{1}y_{2}$
Re($z_{1}$ $z_{2}$ ) = Re$z_{1}$ Re$z_{2}$- Im$z_{1}$ Im$z_{2}$
3. Reduce $\left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$ to the standard form.
= $\left ( \frac{(1+i)-2(1-4i)}{(1-4i)(1+i)} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{1+i-2+8i}{1+i-4i-4i^{2}} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{-1+9i}{1-3i+4} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{-1+9i}{5-3i} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\frac{-3+4i+27i-36i^{2}}{25+5i-15i-3i^{2}}$
= $\frac{-3+31i+36}{25-10i+3}$
= $\frac{33+31i}{28-10i}$
= $\frac{33+31i}{2(14-5i)}$
= $\frac{33+31i}{2(14-5i)}*\frac{14+5i}{14+5i}$
= $\frac{462+i165+i434+155i^{2}}{2(14^{2}-5^{2}i^{2})}$
= $\frac{462+i599-155}{2(196+25)}$
= $\frac{307+i599}{442}$
= $\frac{307}{442}+i\frac{599}{442}$
4. If $x-iy=\sqrt{\frac{a-ib}{c-id}}$ prove that $(x^{2}+y^{2})=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
$x-iy=\sqrt{\frac{a-ib}{c-id}}$
= $\sqrt{\frac{a-ib}{c-id}*\frac{c+id}{c+id}}$
= $\sqrt{\frac{ac+iad-ibc-i^{2}bd}{c^{2}-i^{2}d^{2}}}$
= $\sqrt{\frac{ac+i(ad-bc)+bd}{c^{2}+d^{2}}}$
= $\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}}$
$(x-iy)^{2}$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
$x^{2}+i^{2}y^{2}-i2xy$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
$x^{2}-y^{2}-i2xy$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
On comparing real and imaginary parts we have,
$x^{2}-y^{2}=\frac{ac+bd}{c^{2}+d^{2}}$ and $-2xy=\frac{ad-bc}{c^{2}+d^{2}}$
$(x^{2}+y^{2})^{2}=(x^{2}-y^{2})^{2}+4x^{2}y^{2}$
= $(\frac{ac+bd}{c^{2}+d^{2}})^{2}+(\frac{ad-bc}{c^{2}+d^{2}})^{2}$
= $\frac{a^{2}c^{2}+b^{2}d^{2}+2abcd}{(c^{2}+d^{2})^{2}}$ $+\frac{a^{2}d^{2}+b^{2}c^{2}-2abcd}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}(c^{2}+d^{2})+b^{2}(d^{2}+c^{2})}{(c^{2}+d^{2})^{2}}$
= $\frac{(c^{2}+d^{2})(a^{2}+b^{2})}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
5. Convert the following in the polar form
(i) $\frac{1+i7}{(2-i)^{2}}$
= $\frac{1+i7}{4+i^{2}-i4}$
= $\frac{1+i7}{4-1-i4}$
= $\frac{1+i7}{3-i4}$
= $\frac{1+i7}{3-i4}*\frac{3+i4}{3+i4}$
= $\frac{3+i4+i21+i^{2}28}{3^{2}-i^{2}4^{2}}$
= $\frac{3+i25-28}{9+16}$
= $\frac{-25+i25}{25}$
= $-1+i$
Let $r\cos\theta =-1$ and $r\sin\theta =1$
$r=\sqrt{(-1)^{2}+1^{2}}$
$r=\sqrt{2}$
$\cos \theta=\frac{-1}{r}=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{r}=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Polar form is $z=r(\cos\theta+i\sin\theta)$
= $\sqrt{2}\left [\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}) \right ]$
(ii) $\frac{1+i3}{1-i2}$
= $\frac{1+i3}{1-i2}*\frac{1+i2}{1+i2}$
= $\frac{1+i2+i3+i^{2}6}{1^{2}-i^{2}2^{2}}$
= $\frac{1+i5-6}{1+4}$
= $\frac{-5+i5}{5}$
= $-1+i$
Let $r\cos\theta =-1$ and $r\sin\theta =1$
$r=\sqrt{(-1)^{2}+1^{2}}$
$r=\sqrt{2}$
$\cos \theta=\frac{-1}{r}=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{r}=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Polar form is $z=r(\cos\theta+i\sin\theta)$
= $\sqrt{2}\left [\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}) \right ]$
Solve the following equations
6. $3x^{2}-4x+\frac{20}{3}=0$
$a=3, b=-4, c=\frac{20}{3}$
$b^{2}-4ac=16-80=-64$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{4\pm i\sqrt{64}}{6}$
= $\large\frac{4\pm i8}{6}$
= $\large\frac{2\pm i4}{3}$
7. $x^{2}-2x+\frac{3}{2}=0$
$a=1, b=-2, c=\frac{3}{2}$
$b^{2}-4ac=4-6=-2$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{2\pm i\sqrt{2}}{2}$
8. $27x^{2}-10x+1=0$
$a=27, b=-10, c=1$
$b^{2}-4ac=100-108=-8$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{10\pm i\sqrt{8}}{54}$
= $\large\frac{5\pm i\sqrt{2}}{27}$
9. $21x^{2}-28x+10=0$
$a=21, b=-28, c=10$
$b^{2}-4ac=784-840=-56$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{28\pm i\sqrt{56}}{42}$
= $\large\frac{14\pm i\sqrt{14}}{21}$
10. If $z_{1}=2-i$ and $z_{2}=1+i$, find $|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}|$
$\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}$ $=\frac{2-i+1+i+1}{2-i-1-i+1}$
= $\frac{4}{2-i2}$
= $\frac{2}{1-i}$
= $\frac{2}{1-i}*\frac{1+i}{1+i}$
= $\frac{2+i2}{1^{2}-i^{2}}$
= $\frac{2+i2}{2}$
= $1+i$
$|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}$
11. If $a+ib=\frac{(x+i)^{2}}{2x^{2}+1}$, prove that $a^{2}+b^{2}=\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
$a+ib=\frac{(x+i)^{2}}{2x^{2}+1}$
= $\frac{x^{2}+i^{2}+i2x}{2x^{2}+1}$
= $\frac{x^{2}-1+i2x}{2x^{2}+1}$
= $\frac{x^{2}-1}{2x^{2}+1}+i\frac{2x}{2x^{2}+1}$
On comparing real and imaginary parts we have,
$a=\frac{x^{2}-1}{2x^{2}+1}$ and $b=\frac{2x}{2x^{2}+1}$
$a^{2}+b^{2}$ $=(\frac{x^{2}-1}{2x^{2}+1})^{2}+(\frac{2x}{2x^{2}+1})^{2}$
= $\frac{x^{4}+1-2x^{2}+4x^{2}}{(2x^{2}+1)^{2}}$
= $\frac{x^{4}+1+2x^{2}}{(2x^{2}+1)^{2}}$
= $\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
12. Let $z_{1}=2-i$ and $z_{2}=-2+i$
(i) $\large Re(\frac{z_{1}z_{2}}{\bar{z_{1}}})$
$\frac{z_{1}z_{2}}{\bar{z_{1}}}$ $=\frac{(2-i)(-2+i)}{2+i}$
= $\frac{-4+i2+i2-i^{2}}{2+i}$
= $\frac{-4+i4+1}{2+i}$
= $\frac{-3+i4}{2+i}$
= $\frac{-3+i4}{2+i}*\frac{2-i}{2-i}$
= $\frac{-6+i3+i8-4i^{2}}{4-i^{2}}$
= $\frac{-6+i11+4}{4+1}$
= $\frac{-2+i11}{5}$
$\large Re(\frac{z_{1}z_{2}}{\bar{z_{1}}})=\frac{-2}{5}$
(ii) $\large Im(\frac{1}{z_{1}\bar{z_{1}}})$
$\frac{1}{z_{1}\bar{z_{1}}}$ $=\frac{1}{(2-i)(2+i)}$
= $\frac{1}{2^{2}-i^{2}}$
= $\frac{1}{4+1}=\frac{1}{5}$
$\frac{1}{z_{1}\bar{z_{1}}}=0$
13. Find the modulus and argument of the complex number $\frac{1+i2}{1-i3}$
= $\frac{1+i2}{1-i3}*\frac{1+i3}{1+i3}$
= $\frac{1+i3+i2+i^{2}6}{1^{2}-i^{2}3^{2}}$
= $\frac{1+i5-6}{1+9}$
= $\frac{-5+i5}{10}$
= $\frac{-1+i}{2}$
= $\frac{-1}{2}+i\frac{1}{2}$
modulus r=$\sqrt{\frac{1}{4}+\frac{1}{4}}$
= $\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$
Let $r\cos\theta=-\frac{1}{2}$ and $r\sin\theta=\frac{1}{2}$
$\cos\theta=-\frac{1/2}{1/\sqrt{2}}$ and $\sin\theta=\frac{1/2}{1/\sqrt{2}}$
$\cos\theta=-\frac{1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
14. Find the real numbers x and y if $(x-iy)(3+i5)$ is the conjugate of $-6-i24$
$(x-iy)(3+i5)$ $=3x+i5x-i3y-5yi^{2}$
= $3x+i(5x-3y)+5y$
= $(3x+5y)+i(5x-3y)$
Given that,
$(3x+5y)-i(5x-3y)=-6-i24$
Equating real and imaginry parts we have,
$3x+5y=-6$ ...(i)
$5x-3y=24$ ...(ii)
Solving (i) and (ii) ,
$x=3$ and $y=-3$
15. Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$
$\frac{1+i}{1-i}-\frac{1-i}{1+i}$ $=\frac{(1+i)(1+i)-(1-i)(1-i)}{(1-i)(1+i)}$
= $\frac{1+i^{2}+i2-1-i^{2}+i2}{1-i^{2}}$
= $\frac{i4}{2}=i2$
$|i2|=\sqrt{4}=2$
16. If $(x+iy)^{3}=u+iv$ , then show that $\frac{u}{x}+\frac{v}{y}=4(x^{2}-y^{2})$
$(x+iy)^{3}$ $=x^{3}+i^{3}y^{3}+i3x^{2}y+3x(iy)^{2}$
= $x^{3}-iy^{3}+i3x^{2}y-3xy^{2}$
= $(x^{3}-3xy^{2})+i(3x^{2}y-y^{3})$
∴ $u=x^{3}-3xy^{2}$ and $v=3x^{2}y-y^{3}$
$\frac{u}{x}+\frac{v}{y}$ $=\frac{x^{3}-3xy^{2}}{x}+\frac{3x^{2}y-y^{3}}{y}$
= $x^{2}-3y^{2}+3x^{2}-y^{2}$
= $4x^{2}-4y^{2}$
= $4(x^{2}-y^{2})$
17. If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$, then find $\large |\frac{\beta -\alpha }{1-\bar{\alpha }\beta }|$
Let $\alpha =a+ib$ and $\beta=x+iy$
Given that, $|\beta|=1$
⇒ $\sqrt{x^{2}+y^{2}}=1$
⇒ $x^{2}+y^{2}=1$ ...(i)
$\large \frac{\beta -\alpha }{1-\bar{\alpha }\beta }$ $\large =\frac{x+iy-a-ib}{1-(a-ib)(x+iy)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^{2}by)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+i(ay-bx)+by)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+by+i(ay-bx))}$
= $\large \frac{(x-a)+i(y-b)}{(1-ax+by)+i(ay-bx)}$
$\large |\frac{\beta -\alpha }{1-\bar{\alpha }\beta }|$ $\large =|\frac{(x-a)+i(y-b)}{(1-ax+by)+i(ay-bx)}|$
= $\large \frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-ax+by)^{2}+(ay-bx)^{2}}}$
= $\large \frac{\sqrt{x^{2}+a^{2}-2xa+y^{2}+b^{2}-2yb}}{\sqrt{1-2ax-2by+a^{2}(x^{2}+y^{2})+b^{2}(x^{2}+y^{2})}}$
= $\large \frac{\sqrt{x^{2}+y^{2}+a^{2}-2xa+b^{2}-2yb}}{\sqrt{1-2ax-2by+(x^{2}+y^{2})(a^{2}+b^{2})}}$
= $\large \frac{\sqrt{1+a^{2}+b^{2}-2xa-2by}}{\sqrt{1+a^{2}+b^{2}-2ax-2by}}$
= 1
18. Find the number of non-zero integral solutions of the equation $|1-i|^{x}=2^{x}$
$|1-i|^{x}=2^{x}$
⇒ $(\sqrt{1^{2}+(-1)^{2}})^{x}=2^{x}$
⇒ $(\sqrt{2})^{x}=2^{x}$
⇒ $(2)^{x/2}=2^{x}$
⇒ $\frac{x}{2}=x$
⇒ $x=2x$
⇒ $2x-x=0$
⇒ $x=0$
0 is the only integral solution.
∴ Number of non-zero integral solution is 0.
19. If $(a+ib)(c+id)(e+if)$ $(g+ih)=A+iB$, then show that $(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})$ $(g^{2}+h^{2})=A^{2}+B^{2}$
Given that,
$A+iB$ $=(a+ib)(c+id)(e+if)$ $(g+ih)$
$|A+iB|$ $=|a+ib|$ $*|c+id|$ $*|e+if|$ $*|g+ih|$
$\sqrt{A^{2}+B^{2}}$ $=\sqrt{a^{2}+b^{2}}$ $*\sqrt{c^{2}+d^{2}}$ $*\sqrt{e^{2}+f^{2}}$ $*\sqrt{g^{2}+h^{2}}$
$A^{2}+B^{2}$ $=(a^{2}+b^{2})$ $(c^{2}+d^{2})$ $(e^{2}+f^{2})$ $(g^{2}+h^{2})$
20. If $\left ( \frac{1+i}{1-i} \right )^{m}=1$, then find the least positive integral value of m.
$\left ( \frac{1+i}{1-i} \right )^{m}=1$
$\left ( \frac{1+i}{1-i}* \frac{1+i}{1+i}\right )^{m}=1$
$\left ( \frac{(1+i)^{2}}{1^{2}-i^{2}}\right )^{m}=1$
$\left ( \frac{1^{2}+i^{2}+2i}{1+1}\right )^{m}=1$
$\left ( \frac{1-1+i2}{2}\right )^{m}=1$
$i^{m}=1$
$i^{m}=i^{4k}$
$m=4k$
where k is some integer with least value 1.
∴ the least positive integral value of m is 4.
= $\left [ (i^{4})^{4}.i^{2}+\frac{1}{(i^{4})^{6}.i} \right ]^{3}$
= $\left [ (1).(-1)+\frac{1}{(1).i} \right ]^{3}$
= $\left [ -1+\frac{1}{i}*\frac{i}{i} \right ]^{3}$
= $\left [ -1+\frac{i}{i^{2}} \right ]^{3}$
= $\left [ -1+\frac{i}{-1} \right ]^{3}$
= $\left [ -1-i \right ]^{3}$
= $(-1)^{3}(1+i)^{3}$
= $-(1+3i+3i^{2}+i^{3})$
= $-(1+3i-3-i)$
= $-(-2+i2)$
= $2-i2$
2. For any two complex numbers $z_{1}$ and $z_{2}$ prove that Re($z_{1}$ $z_{2}$ ) = Re$z_{1}$ Re$z_{2}$- Im$z_{1}$ Im$z_{2}$
Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$
$z_{1}z_{2}$ $=(x_{1}+iy_{1})(x_{2}+iy_{2})$
= $x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x_{2}+i^{2}y_{1}y_{2}$
= $x_{1}x_{2}+i(x_{1}y_{2}+y_{1}x_{2})-y_{1}y_{2}$
= $(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})$
Re($z_{1}$ $z_{2}$ )$=x_{1}x_{2}-y_{1}y_{2}$
Re($z_{1}$ $z_{2}$ ) = Re$z_{1}$ Re$z_{2}$- Im$z_{1}$ Im$z_{2}$
3. Reduce $\left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )$ to the standard form.
= $\left ( \frac{(1+i)-2(1-4i)}{(1-4i)(1+i)} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{1+i-2+8i}{1+i-4i-4i^{2}} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{-1+9i}{1-3i+4} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\left ( \frac{-1+9i}{5-3i} \right )\left ( \frac{3-4i}{5+i} \right )$
= $\frac{-3+4i+27i-36i^{2}}{25+5i-15i-3i^{2}}$
= $\frac{-3+31i+36}{25-10i+3}$
= $\frac{33+31i}{28-10i}$
= $\frac{33+31i}{2(14-5i)}$
= $\frac{33+31i}{2(14-5i)}*\frac{14+5i}{14+5i}$
= $\frac{462+i165+i434+155i^{2}}{2(14^{2}-5^{2}i^{2})}$
= $\frac{462+i599-155}{2(196+25)}$
= $\frac{307+i599}{442}$
= $\frac{307}{442}+i\frac{599}{442}$
4. If $x-iy=\sqrt{\frac{a-ib}{c-id}}$ prove that $(x^{2}+y^{2})=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
$x-iy=\sqrt{\frac{a-ib}{c-id}}$
= $\sqrt{\frac{a-ib}{c-id}*\frac{c+id}{c+id}}$
= $\sqrt{\frac{ac+iad-ibc-i^{2}bd}{c^{2}-i^{2}d^{2}}}$
= $\sqrt{\frac{ac+i(ad-bc)+bd}{c^{2}+d^{2}}}$
= $\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^{2}+d^{2}}}$
$(x-iy)^{2}$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
$x^{2}+i^{2}y^{2}-i2xy$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
$x^{2}-y^{2}-i2xy$ $=\frac{ac+bd}{c^{2}+d^{2}}+i\frac{ad-bc}{c^{2}+d^{2}}$
On comparing real and imaginary parts we have,
$x^{2}-y^{2}=\frac{ac+bd}{c^{2}+d^{2}}$ and $-2xy=\frac{ad-bc}{c^{2}+d^{2}}$
$(x^{2}+y^{2})^{2}=(x^{2}-y^{2})^{2}+4x^{2}y^{2}$
= $(\frac{ac+bd}{c^{2}+d^{2}})^{2}+(\frac{ad-bc}{c^{2}+d^{2}})^{2}$
= $\frac{a^{2}c^{2}+b^{2}d^{2}+2abcd}{(c^{2}+d^{2})^{2}}$ $+\frac{a^{2}d^{2}+b^{2}c^{2}-2abcd}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}(c^{2}+d^{2})+b^{2}(d^{2}+c^{2})}{(c^{2}+d^{2})^{2}}$
= $\frac{(c^{2}+d^{2})(a^{2}+b^{2})}{(c^{2}+d^{2})^{2}}$
= $\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$
5. Convert the following in the polar form
(i) $\frac{1+i7}{(2-i)^{2}}$
= $\frac{1+i7}{4+i^{2}-i4}$
= $\frac{1+i7}{4-1-i4}$
= $\frac{1+i7}{3-i4}$
= $\frac{1+i7}{3-i4}*\frac{3+i4}{3+i4}$
= $\frac{3+i4+i21+i^{2}28}{3^{2}-i^{2}4^{2}}$
= $\frac{3+i25-28}{9+16}$
= $\frac{-25+i25}{25}$
= $-1+i$
Let $r\cos\theta =-1$ and $r\sin\theta =1$
$r=\sqrt{(-1)^{2}+1^{2}}$
$r=\sqrt{2}$
$\cos \theta=\frac{-1}{r}=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{r}=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Polar form is $z=r(\cos\theta+i\sin\theta)$
= $\sqrt{2}\left [\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}) \right ]$
(ii) $\frac{1+i3}{1-i2}$
= $\frac{1+i3}{1-i2}*\frac{1+i2}{1+i2}$
= $\frac{1+i2+i3+i^{2}6}{1^{2}-i^{2}2^{2}}$
= $\frac{1+i5-6}{1+4}$
= $\frac{-5+i5}{5}$
= $-1+i$
Let $r\cos\theta =-1$ and $r\sin\theta =1$
$r=\sqrt{(-1)^{2}+1^{2}}$
$r=\sqrt{2}$
$\cos \theta=\frac{-1}{r}=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{r}=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
Polar form is $z=r(\cos\theta+i\sin\theta)$
= $\sqrt{2}\left [\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}) \right ]$
Solve the following equations
6. $3x^{2}-4x+\frac{20}{3}=0$
$a=3, b=-4, c=\frac{20}{3}$
$b^{2}-4ac=16-80=-64$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{4\pm i\sqrt{64}}{6}$
= $\large\frac{4\pm i8}{6}$
= $\large\frac{2\pm i4}{3}$
7. $x^{2}-2x+\frac{3}{2}=0$
$a=1, b=-2, c=\frac{3}{2}$
$b^{2}-4ac=4-6=-2$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{2\pm i\sqrt{2}}{2}$
8. $27x^{2}-10x+1=0$
$a=27, b=-10, c=1$
$b^{2}-4ac=100-108=-8$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{10\pm i\sqrt{8}}{54}$
= $\large\frac{5\pm i\sqrt{2}}{27}$
9. $21x^{2}-28x+10=0$
$a=21, b=-28, c=10$
$b^{2}-4ac=784-840=-56$
$x=\frac{-b\pm i\sqrt{4ac-b^{2}}}{2a}$
= $\large\frac{28\pm i\sqrt{56}}{42}$
= $\large\frac{14\pm i\sqrt{14}}{21}$
10. If $z_{1}=2-i$ and $z_{2}=1+i$, find $|\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}|$
$\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}$ $=\frac{2-i+1+i+1}{2-i-1-i+1}$
= $\frac{4}{2-i2}$
= $\frac{2}{1-i}$
= $\frac{2}{1-i}*\frac{1+i}{1+i}$
= $\frac{2+i2}{1^{2}-i^{2}}$
= $\frac{2+i2}{2}$
= $1+i$
$|1+i|=\sqrt{1^{2}+1^{2}}=\sqrt{2}$
11. If $a+ib=\frac{(x+i)^{2}}{2x^{2}+1}$, prove that $a^{2}+b^{2}=\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
$a+ib=\frac{(x+i)^{2}}{2x^{2}+1}$
= $\frac{x^{2}+i^{2}+i2x}{2x^{2}+1}$
= $\frac{x^{2}-1+i2x}{2x^{2}+1}$
= $\frac{x^{2}-1}{2x^{2}+1}+i\frac{2x}{2x^{2}+1}$
On comparing real and imaginary parts we have,
$a=\frac{x^{2}-1}{2x^{2}+1}$ and $b=\frac{2x}{2x^{2}+1}$
$a^{2}+b^{2}$ $=(\frac{x^{2}-1}{2x^{2}+1})^{2}+(\frac{2x}{2x^{2}+1})^{2}$
= $\frac{x^{4}+1-2x^{2}+4x^{2}}{(2x^{2}+1)^{2}}$
= $\frac{x^{4}+1+2x^{2}}{(2x^{2}+1)^{2}}$
= $\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$
12. Let $z_{1}=2-i$ and $z_{2}=-2+i$
(i) $\large Re(\frac{z_{1}z_{2}}{\bar{z_{1}}})$
$\frac{z_{1}z_{2}}{\bar{z_{1}}}$ $=\frac{(2-i)(-2+i)}{2+i}$
= $\frac{-4+i2+i2-i^{2}}{2+i}$
= $\frac{-4+i4+1}{2+i}$
= $\frac{-3+i4}{2+i}$
= $\frac{-3+i4}{2+i}*\frac{2-i}{2-i}$
= $\frac{-6+i3+i8-4i^{2}}{4-i^{2}}$
= $\frac{-6+i11+4}{4+1}$
= $\frac{-2+i11}{5}$
$\large Re(\frac{z_{1}z_{2}}{\bar{z_{1}}})=\frac{-2}{5}$
(ii) $\large Im(\frac{1}{z_{1}\bar{z_{1}}})$
$\frac{1}{z_{1}\bar{z_{1}}}$ $=\frac{1}{(2-i)(2+i)}$
= $\frac{1}{2^{2}-i^{2}}$
= $\frac{1}{4+1}=\frac{1}{5}$
$\frac{1}{z_{1}\bar{z_{1}}}=0$
13. Find the modulus and argument of the complex number $\frac{1+i2}{1-i3}$
= $\frac{1+i2}{1-i3}*\frac{1+i3}{1+i3}$
= $\frac{1+i3+i2+i^{2}6}{1^{2}-i^{2}3^{2}}$
= $\frac{1+i5-6}{1+9}$
= $\frac{-5+i5}{10}$
= $\frac{-1+i}{2}$
= $\frac{-1}{2}+i\frac{1}{2}$
modulus r=$\sqrt{\frac{1}{4}+\frac{1}{4}}$
= $\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}$
Let $r\cos\theta=-\frac{1}{2}$ and $r\sin\theta=\frac{1}{2}$
$\cos\theta=-\frac{1/2}{1/\sqrt{2}}$ and $\sin\theta=\frac{1/2}{1/\sqrt{2}}$
$\cos\theta=-\frac{1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{\sqrt{2}}$
$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$
14. Find the real numbers x and y if $(x-iy)(3+i5)$ is the conjugate of $-6-i24$
$(x-iy)(3+i5)$ $=3x+i5x-i3y-5yi^{2}$
= $3x+i(5x-3y)+5y$
= $(3x+5y)+i(5x-3y)$
Given that,
$(3x+5y)-i(5x-3y)=-6-i24$
Equating real and imaginry parts we have,
$3x+5y=-6$ ...(i)
$5x-3y=24$ ...(ii)
Solving (i) and (ii) ,
$x=3$ and $y=-3$
15. Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$
$\frac{1+i}{1-i}-\frac{1-i}{1+i}$ $=\frac{(1+i)(1+i)-(1-i)(1-i)}{(1-i)(1+i)}$
= $\frac{1+i^{2}+i2-1-i^{2}+i2}{1-i^{2}}$
= $\frac{i4}{2}=i2$
$|i2|=\sqrt{4}=2$
16. If $(x+iy)^{3}=u+iv$ , then show that $\frac{u}{x}+\frac{v}{y}=4(x^{2}-y^{2})$
$(x+iy)^{3}$ $=x^{3}+i^{3}y^{3}+i3x^{2}y+3x(iy)^{2}$
= $x^{3}-iy^{3}+i3x^{2}y-3xy^{2}$
= $(x^{3}-3xy^{2})+i(3x^{2}y-y^{3})$
∴ $u=x^{3}-3xy^{2}$ and $v=3x^{2}y-y^{3}$
$\frac{u}{x}+\frac{v}{y}$ $=\frac{x^{3}-3xy^{2}}{x}+\frac{3x^{2}y-y^{3}}{y}$
= $x^{2}-3y^{2}+3x^{2}-y^{2}$
= $4x^{2}-4y^{2}$
= $4(x^{2}-y^{2})$
17. If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$, then find $\large |\frac{\beta -\alpha }{1-\bar{\alpha }\beta }|$
Let $\alpha =a+ib$ and $\beta=x+iy$
Given that, $|\beta|=1$
⇒ $\sqrt{x^{2}+y^{2}}=1$
⇒ $x^{2}+y^{2}=1$ ...(i)
$\large \frac{\beta -\alpha }{1-\bar{\alpha }\beta }$ $\large =\frac{x+iy-a-ib}{1-(a-ib)(x+iy)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^{2}by)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+i(ay-bx)+by)}$
= $\large \frac{(x-a)+i(y-b)}{1-(ax+by+i(ay-bx))}$
= $\large \frac{(x-a)+i(y-b)}{(1-ax+by)+i(ay-bx)}$
$\large |\frac{\beta -\alpha }{1-\bar{\alpha }\beta }|$ $\large =|\frac{(x-a)+i(y-b)}{(1-ax+by)+i(ay-bx)}|$
= $\large \frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-ax+by)^{2}+(ay-bx)^{2}}}$
= $\large \frac{\sqrt{x^{2}+a^{2}-2xa+y^{2}+b^{2}-2yb}}{\sqrt{1-2ax-2by+a^{2}(x^{2}+y^{2})+b^{2}(x^{2}+y^{2})}}$
= $\large \frac{\sqrt{x^{2}+y^{2}+a^{2}-2xa+b^{2}-2yb}}{\sqrt{1-2ax-2by+(x^{2}+y^{2})(a^{2}+b^{2})}}$
= $\large \frac{\sqrt{1+a^{2}+b^{2}-2xa-2by}}{\sqrt{1+a^{2}+b^{2}-2ax-2by}}$
= 1
18. Find the number of non-zero integral solutions of the equation $|1-i|^{x}=2^{x}$
$|1-i|^{x}=2^{x}$
⇒ $(\sqrt{1^{2}+(-1)^{2}})^{x}=2^{x}$
⇒ $(\sqrt{2})^{x}=2^{x}$
⇒ $(2)^{x/2}=2^{x}$
⇒ $\frac{x}{2}=x$
⇒ $x=2x$
⇒ $2x-x=0$
⇒ $x=0$
0 is the only integral solution.
∴ Number of non-zero integral solution is 0.
19. If $(a+ib)(c+id)(e+if)$ $(g+ih)=A+iB$, then show that $(a^{2}+b^{2})(c^{2}+d^{2})(e^{2}+f^{2})$ $(g^{2}+h^{2})=A^{2}+B^{2}$
Given that,
$A+iB$ $=(a+ib)(c+id)(e+if)$ $(g+ih)$
$|A+iB|$ $=|a+ib|$ $*|c+id|$ $*|e+if|$ $*|g+ih|$
$\sqrt{A^{2}+B^{2}}$ $=\sqrt{a^{2}+b^{2}}$ $*\sqrt{c^{2}+d^{2}}$ $*\sqrt{e^{2}+f^{2}}$ $*\sqrt{g^{2}+h^{2}}$
$A^{2}+B^{2}$ $=(a^{2}+b^{2})$ $(c^{2}+d^{2})$ $(e^{2}+f^{2})$ $(g^{2}+h^{2})$
20. If $\left ( \frac{1+i}{1-i} \right )^{m}=1$, then find the least positive integral value of m.
$\left ( \frac{1+i}{1-i} \right )^{m}=1$
$\left ( \frac{1+i}{1-i}* \frac{1+i}{1+i}\right )^{m}=1$
$\left ( \frac{(1+i)^{2}}{1^{2}-i^{2}}\right )^{m}=1$
$\left ( \frac{1^{2}+i^{2}+2i}{1+1}\right )^{m}=1$
$\left ( \frac{1-1+i2}{2}\right )^{m}=1$
$i^{m}=1$
$i^{m}=i^{4k}$
$m=4k$
where k is some integer with least value 1.
∴ the least positive integral value of m is 4.
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