Question 3
Express the following complex number in the standard form
(i) $(1 + 2i)( 4 + 3i)$
(ii) $(1 + i)^6 + (1 – i)^3$
(iii) $(2 +i)^4$
(iv)$\frac {(3+4i)(4 + 3i)}{1+i}$ Solution
Question 4
Find the polar form of the following complex numbers
(i) (1 + i)
(ii) $\frac {1 + i}{1-i}$
(iii) 5
(iv) $2\frac { i-1}{1 + i \sqrt 3}$ Solution
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
$\sqrt {2} (cos \frac {5 \pi}{12} + i sin \frac {5 \pi}{12})$
Question 5
Find the square roots of the following complex numbers
(a) 5 -2i
(b) -24 -18i Solution
(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
$\sqrt {-24 -18i} = \pm \sqrt {3}( 1-3i)$
Question 6
True and False
(i) Conjugate of 1+i lies in IV Quadrant
(ii) Polar form of the complex number $(i^{25})^3$ is $cos \frac {\pi}{2} -i sin \frac {\pi}{2}$
(iii) Principle argument for $z=-10$ is $-\pi$
(iv) The order relation is defined on the set of complex numbers.
(v)5 is a complex number
(vi) If $|z_1| =|z_2|$ then $z_1=z_2$
(vii) if |z|=5, then locus of the complex number is a circle
(viii) $|z_1 + z_2 + z_3 ...+ z_n| \leq |z_1| + |z_2| + |z_3| +.... + |z_n|$ Solution