Question

complex number exercise my tentative with picture?

Answer 1

I try to slove this exercise similar in metodfriend but I failed. Waht I do wrong?

Answer 2

@mathmath333 can you help please?

Answer 3

@Directrix are you available to help?

Answer 4

you maen me?

Answer 5

@BPDlkeme234 Happy New Year do you mind helping?

Answer 6

whats ur exact question @franzmller682

Answer 7

I will slove this exercise on the first picture like the second picture.

Answer 8

@mathmath333 thank you

Answer 9

also my tentativ is the same in the second picture the I trasform kordinate in porlar

Answer 10

i post the font of exercise

Answer 11

i think deMoivre's rule is what we need

Answer 12

oh but you don't actually need to simplify the 4000

Answer 13

ok I take 5 or 10 minutes to post the font of exercise is on smartphone

Answer 14

i think your r is wrong

Answer 15

formula
\(\large\tt \begin{align} \color{black}{
(\cos x+i \sin x)^n=\cos (nx)+i \sin (nx)
}\end{align}\)

Answer 16

shouldn't be negative after squaring, so r=sqrt2

Answer 17

let -1=rcos \[-1=r \cos \theta,-1\iota=\iota r \sin \theta ,r \cos \theta=-1,r \sin \theta=-1\]
square and add
\[r^2\left( \cos ^2\theta +\sin ^2\theta \right)=\left( -1 \right)^2+\left( -1 \right)^2=2,r=\sqrt{2}\]

Answer 18

this is german but I can translate the solution is in geramn lösung

Answer 19

Ok the first I understand but When I made tangens the solution is then pi/4 but this is wrong

Answer 20

you need to consider that all the tangent repeats, and tan(pi/4)=tan(5pi/4)

Answer 21

because both components are negative you are in quadrant III, not quadrant I, so the angle is greater than pi

Answer 22

\(\frac xy=\frac{-x}{-y}\) so \(\tan(\frac xy)=\tan(\frac{-x}{-y})\)

Answer 23

and \[\tan^{-1}u=\{\frac xy,\frac{-x}{-y}\}\]you always need to think of the unit circle to decide the right angle

Answer 24

Ok in this case the tangen have tan^-1(-1/-1) but then is 1 grade and one grade is then pi/4

Answer 25

arctan has two values associated with every argument because y/x=-y/-x

Answer 26

sin and cos both are negative so angle should be in third quadrant

\[-1=\sqrt{2}\cos \theta,\cos \theta=\frac{ -1 }{ \sqrt{2} }=-\cos \frac{ \pi }{ 4 }=\cos \left( \pi+\frac{ \pi }{ 4 } \right)\]
\[\theta=\frac{ 5 \pi }{ 4 }\]

Answer 27

in fact adding \(\pi\) will always keep the tan the same

Answer 28

\[\tan^{-1}1=\frac\pi4+n\pi,~n\in\mathbb Z\]

Answer 29

Ok thx I take time to understand it.

Answer 30

\[z=(-1-j)^{4000} ~~solution ~~ z=(\sqrt{2}e ^{j5/4\pi})^{4000}\]

Answer 31

now when I do tagens.

Answer 32

\[\tan ^{-1}(\frac{ -1 }{ -1 })=1 ~and~ 1 ~is ~then~ 45 ~grade~ or ~is ~false ~I ~study ~unit ~circle ~but ~I ~understad ~a ~littlet~ bit ~\]

Answer 33

but I understand it not so

Answer 34

your calculator will only give you one value for each trigonometric function, but in reality each each trig function has infinite answers on two places on the unit circle

have you taken trigonometry yet?
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf

Answer 35

I must refreh it

Answer 36

indeed, look at other functions like sine and cosine. see how sin(pi/4)=sin(3pi/4)
?

Answer 37

my font is this:

Answer 38

I can't understand how I can arrive of 5/4 pi

Answer 39

look, what is the definition of the tangent?

Answer 40

tan=sin/cos

Answer 41

yes, or simply opposite/adjacent, or y/x
right?

Answer 42

yes

Answer 43

what is the tangent of this angle?