Question

what changes we make in different quadrants to find argument..give the difference
1)Convert the given complex number in polar form: 1 – i
2)Convert the given complex number in polar form: – 1 + i

Answer 1

@phi

Answer 2

do you know trigonometry?

Answer 3

yes ofcourse

Answer 4

plot 1 – i
and find the angle (measured from the x-axis, going counter-clockwise)

Answer 5

next, find the length of the hypotenuse
in this case sqr(2)

now write
\[ Ae^{i\theta} \]
where A is the length and theta is the angle

Answer 6

people generally write the angle in radians

Answer 7

\[ 1 – i = \sqrt{2}e^{-i \frac{\pi}{4}} \]

\[ -1 + i = \sqrt{2}e^{i \frac{3\pi}{4}} \]
notice if we add pi to the first number's angle, we get 3pi/4
and if we look at the plot we see the two points form a line.

Answer 8

\[\cos \theta=1/\sqrt{2} and \sin \theta=-1/\sqrt{2}\]
i have got this so far now i want to know it further

Answer 9

@phi

Answer 10

@goformit100

Answer 11

I posted the result up above
\[ 1 – i = \sqrt{2}e^{-i \frac{\pi}{4}} = \sqrt{2}\left(\cos\left( -\frac{\pi}{4}\right) + i \sin\left( -\frac{\pi}{4}\right)\right) \]
using trig identities we can write that as
\[ \sqrt{2}\left(\cos\left( \frac{\pi}{4}\right) - i \sin\left( \frac{\pi}{4}\right)\right) \]

note that if we sub in the numeric values for cos and sin we get
\[ \sqrt{2}\left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}}\right) = 1 - i\]
and we recover the rectangular form

Answer 12

cant it be like sintheta/costheta=tantheta

Answer 13

coz we are taught like this and it is acceptable only

Answer 14

\large \theta = \tan^{-1}(-1) = -\dfrac{\pi}{4}

Answer 15

sin(x)/cos(x) = tan(x) is an identity

Do you have a specific question?

Perhaps you are using a different notation for the polar form?
These all mean the same thing
\[ r \ cis \ \theta \\ r \angle \theta \\ r e^{i \theta} \\ r(\cos \theta + i \sin \theta)\]

Answer 16

to find theta associated with complex number x you would use
\[ \tan^{-1}\left( \frac{imag(x)}{real(x)}\right) \]

Answer 17

like we got r=\[\sqrt{2}\]

Answer 18

http://www.meritnation.com/cbse/class11-science/ncert-solutions/math/math---ncert-solution/complex-numbers-and-quadratic-equations/page108-exercise-5-2-qno3/1342_165_4136/convert-the-given-complex-number-in-polar-form-1-ndash-i

Answer 19

I would use pythagorean theorem to find r

\[ r = \sqrt{real(x))^2 + imag(x)^2} \]

Answer 20

\[i use r(costheta+isintheta)\]

Answer 21

then the answer can be written as
\[\sqrt{2}\left(\cos\left( -\frac{\pi}{4}\right) + i \sin\left( -\frac{\pi}{4}\right)\right) \\
\sqrt{2}\left(\cos\left( \frac{\pi}{4}\right) - i \sin\left( \frac{\pi}{4}\right)\right)\\
\sqrt{2}\cos\left( \frac{\pi}{4}\right) - i \sqrt{2}\sin\left( \frac{\pi}{4}\right)
\]

Answer 22

but how do we get pi/4 ????????

Answer 23

or
\[ \sqrt{2} \left( \cos(\frac{7 \pi}{4}) + i \sin(\frac{7 \pi}{4}) \right)\]