Complex Numbers Worksheet

Multiplicative inverse is given by
$\frac{1}{z}=\frac{\bar{z}}{|z|^{2}}$
(i) $\frac {3}{16} + \frac { \sqrt 7}{16} i$

(ii) $\frac {2}{13} - \frac {3}{13} i$

(iii) $(1 + \sqrt {3} i)^2 = 1 - 3 + 2 \sqrt {3} i= -2 + 2 \sqrt {3} i$
Multiplicative inverse will be
$\frac {-2}{16} - \frac {2 \sqrt 3}{16} i = \frac {-1}{8} - \frac {\sqrt 3}{8} i$

(iv) $ \frac {3}{25} - \frac {4}{25} i$

(i) $1 + \sqrt {2} i$

(ii) 5 + i

(iii) $\frac {1}{3 - 4i} = \frac {1}{3 - 4i} \times \frac {3 + 4i}{3 + 4i}= \frac { 3+ 4i}{3^2 - (4i)^2} = \frac {3}{25} + \frac {4}{25} i$
conjugate will be
$\frac {3}{25} - \frac {4}{25} i$

(iv) $\frac {1+i}{1-i} = \frac {1+i}{1-i} \times \frac {1+i}{1+i}= \frac {1 + i^2 + 2i}{1 -i^2} = \frac {2i}{2} = i$
conjugate will be $-i$
(v)$\frac { \sqrt {3 + 12 i} + \sqrt {3 - 12 i}}{\sqrt {3 + 12 i} - \sqrt {3 - 12 i}} = \frac { \sqrt {3 + 12 i} + \sqrt {3 - 12 i}}{\sqrt {3 + 12 i} - \sqrt {3 - 12 i}} \times \frac { \sqrt {3 + 12 i} + \sqrt {3 - 12 i}}{\sqrt {3 + 12 i} + \sqrt {3 - 12 i}}= \frac {(\sqrt {3 + 12 i} + \sqrt {3 - 12 i})^2}{3+ 12i -(3 -12i)}$ $=\frac {3+ 12i + 3 -12i + 2 \sqrt {3^2 -(12i)^2}}{24i} = \frac {6 + 2\sqrt {153}}{24i}= \frac { 3 + \sqrt {153}}{12i} = - \frac {3 + \sqrt {153}}{12} i$
Conjugate will be
$ 0 + \frac {3 + \sqrt {153}}{12} i$

(i) $ 4 + 3i+ 8i + 6i^2 = -2 + 9i$

(ii)$(1 + i)^6= {(1 + i)^2}^3 = (1 + i^2+ 2i)^3= (1 – 1 + 2i)^3= 8i^3= – 8i$
and $(1 – i)^3= 1 – i^3– 3i + 3i^2= 1 + i – 3i – 3 = – 2 – 2i$
Therefore, $(1 + i)^6 + (1 – i)^3$
$= – 8i – 2 – 2i = – 2 – 10i$

Steps to obtained the Polar Form
(1) First express it in the form a + ib
(2) Now $r = \sqrt {a^2 + b^2}$
(3) The principal argument of the complex number is find using the below steps
(i) find the acute angle value of $\theta=tan^{-1}|\frac{y}{x}|$
(ii)Look for the values of a ,b
if (a,b) lies in First quadrant then Argument=$\theta$
if (a,b) lies in second quadrant then Argument =$\pi-\theta$
if (a,b) lies in third quadrant then Argument =$-\pi+\theta$
if (a,b) lies in Fourth quadrant then Argument =$-\theta$
(4) Polar Form is written as
$r[cos \theta + isin \theta]$
(i) $(1 + i) = \sqrt {2} (cos \frac {\pi}{4} + i cos \frac {\pi}{4})$

(ii) $\frac {1 + i}{1-i} = \frac {1 + i}{1-i} \times \frac {1 + i}{1+i}= \frac {1 -2i + i^2}{1-i^2} = i$
$r= \sqrt {0 + 1}=1$
Now point (0,1) lies on Positive y -axis
$\theta = \frac {\pi}{2}$
Hence polar form
$1(cos \frac {\pi}{2} + i sin \frac {\pi}{2})$

(iii) $5 = 5 + i (0)$
$r= \sqrt {5^ + 0}=5$
Now point (5,0) lies on Positive x -axis
$\theta = 0$
Hence polar form
$5(cos 0 + i sin 0$

(iv) $2\frac { i-1}{1 + i \sqrt 3} = 2\frac { i-1}{1 + i \sqrt 3} \times \frac { 1 - i \sqrt 3}{1 - i \sqrt 3}= 2 \frac {i -i^2 \sqrt 3 -1+ i \sqrt {3}}{1+3}= \frac { \sqrt 3 -1}{2} + i \frac { \sqrt 3 + 1}{2}$
$r = \sqrt { (\frac { \sqrt 3 -1}{2})^2 + (\frac { \sqrt 3 + 1}{2})^2 } = \sqrt 2$
Now
$r cos \theta = \frac { \sqrt 3 -1}{2}$
$cos \theta = \frac { \sqrt 3 -1}{2 \sqrt 2}= \frac {\sqrt 3}{2} \frac {1}{\sqrt 2} - \frac {1}{2} \frac {1}{\sqrt 2} = cos \frac {\pi}{3} cos \frac {\pi}{4} - sin \frac {\pi}{3} sin \frac {\pi}{4}=cos (\frac {\pi}{3} + \frac {\pi}{4})$
So, $\theta = \frac {\pi}{3} + \frac {\pi}{4}= \frac {5 \pi}{12}$
Hence polar form
$\sqrt {2} (cos \frac {5 \pi}{12} + i sin \frac {5 \pi}{12})$

(a) Let $ \sqrt {5- 2i}=x+iy$
Squaring both the sides
$5- 2i=(x+iy)^2$
Then, $x^2- y^2=5$, $2xy= -12$ -(1)
Now
$(x^2 + y^2)^2 = (x^2 -y^2)^2 + 4x^2y^2$
$(x^2 + y^2)^2 =(5)^2 + (-12)^2$
With no ambiguity of sign, since $x^2 + y^2 > O$.
$x^2 + y^2=13$ --(2)
It now follows from (1) and (2)
$2x^2=5+13=18$
$x= \pm 3$
and then
$y= -\frac {6}{x}=\mp 2$
Hence
$\sqrt {5 -2i}= \pm (3- 2i)$
(b) This can also be similary as above
$\sqrt {-24 -18i} = \pm \sqrt {3}( 1-3i)$

(i) True
(ii) True
(iii) False
(iv) False
(v) True
(vi) false
(vii) True
(viii) True

$(x+ iy)^{1/3} =a + ib$
$x + iy=(a+ ib)^3$
$x+ iy = a^3 + i^3b^3 + 3a^2ib + 3ai^2b^2$
$x+ iy= (a^3 -3ab^2) + i(3a^2b -b^3)$
So,
$x= a^3 -3ab^2$ and $y=3a^2b -b^3$
$\frac {x}{a}= a^2 -3b^2$ and $\frac {y}{b}=3a^2 -b^2$
Adding both
$\frac {x}{a} + \frac {y}{b}= 4(a^2 -b^2)$

Let $z = x + iy$, $z_1= x_1 + iy_1$ and $z_2= x_2 + iy_2$
Now as per the question
$z + \bar {z} = 2|z-1|$
Therefore
$(x + iy) + (x – iy) = 2 |x -1 +iy|$
or $2x = 1 + y^2$
Similary , we have
$2x_1 = 1 + y_1^2$ and $2x_2 = 1 + y_2^2$
Or
$2(x_1 -x_2) = y_1^2 - y_2^2$
$2(x_1-x_2) = (y_1 - y_2)(y_1 + y_2)$ -(1)
Now
$z_1 - z_2 = x_1 - x_2 + i(y_1 -y_2)$
$arg (z_1 -z_2) = tan ^{-1} \frac {y_1 - y_2}{x_1 - x_2}$
$\frac {\pi}{4} = tan ^{-1} \frac {y_1 - y_2}{x_1 - x_2}$
or $\frac {y_1 - y_2}{x_1 - x_2}=1$ -(2)
From (1) and (2)
$y_1 + y_2 = 2$
So, $Im (z_1 + z_2)=2$