Question

arg(5-√3i). help for the answer
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Answer 1

Step-by-step explanation:

0) If

z cos isin

6 6

 

 

then

A)

|z| 1, arg z

6



 

B)

3 5 |z| , arg z

2 24



 

C) | z | 1, arg z

4



 

D)

3 1 1 |z| , arg z tan

Answer 2

Answer: Argument of $(5-\sqrt{3}i )$  is  $-2\pi +cos^{-1} \frac{5}{2\sqrt{7} }$

Step-by-step explanation:

 arg (z) = $(5-\sqrt{3}i )$

complex number z is of the form $x+iy$,

by comparing the value of z we can say that ,

the value of x = 5 ,

the value of y = $-\sqrt{3}$,

so,

the modulus of z  is

=>  |z| = $\sqrt{x^{2} +y^{2} }$

=>    |z| = $\sqrt{5^{2}+(-\sqrt{3} ^{2}) }$

=>    |z| = $\sqrt{25+3}$

=>    |z| = $\sqrt{28}$

=>    |z| = $2\sqrt{7}$

so , r =  $2\sqrt{7}$

now,

       z = r ( cosθ + isinθ)

       $(5-\sqrt{3}i )$  =  r ( cosθ + isinθ)

   

now, comparing the real parts of the equation,

=> 5 = r cosθ

=> 5 = $2\sqrt{7}$ cosθ

=> cosθ = $\frac{5}{2\sqrt{7} }$

now , comparing the imaginary parts of the equation,

=> $-\sqrt{3}i$ = r isinθ

=> $-\sqrt{3}$  = r sinθ

=> $-\sqrt{3}$  = $2\sqrt{7}$ sinθ

=> sinθ  = $\frac{-\sqrt{3}}{2\sqrt{7} }$

sinθ is negative and cosθ is positive ,

it is possible only in IV quadrant .

so,

Argument = $-(360^{0} - cos^{-1} \frac{5}{2\sqrt{7} } )$

                   =>  $-360^{0} + cos^{-1} \frac{5}{2\sqrt{7} }$

                    => $-2\pi +cos^{-1} \frac{5}{2\sqrt{7} }$

so, the argument of $(5-\sqrt{3}i )$  is equal to $-2\pi +cos^{-1} \frac{5}{2\sqrt{7} }$

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