help me with this! (1 Viewer)

[muttiah]

Find the modulus and argument and express in the form r(cost + isint)

2[-cospi/3 + isin pi/3)


?

 

[SoulSearcher]

-cos(pi/3) + isin(pi/3) = cos(2pi/3) + isin(2pi/3)
Therefore |z| = 2 and Arg z = 2pi/3
Therefore it is 2(cos(2pi/3) + isin(2pi/3))

 

[muttiah]

any other longer ways?

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if z = (1-2i)(i-3) find the a)multiplicative inverse b)additive inverse

what does that mean?

 

[SoulSearcher]

Multiplcative inverse, conjugate that when multiplied with the original gives an answer of 1.

Additive inverse, inverse that when added to the original gives the value of 0.

 

[muttiah]

kk thanks got it

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|re(z)| < |z|

let z = x+iy

sqrtx < sqrtx+iy

x< x+iy

how can i prove this?

 

[jyu]

kk thanks got it

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|re(z)| < |z|

let z = x+iy

sqrtx < sqrtx+iy

x< x+iy

how can i prove this?

x^2 < x^2 + y^2 for y =/= 0,
|x|^2 < |x+iy|^2,
|x| < |x+iy|,
hence |re(z)| < |z|, where z = x+iy.

 

[muttiah]

kk cool 2 methods..

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|z-i|=|z+2-3i|

let z = x+iy

|x+iy-i| = |x+iy+2-3i|

|x^2+y^2-i|=|x^2+y^2+2-3i|

x^2+y^2 - i = x^2 +y^2 + 2 -3i

and then?

 

[pLuvia]

I'm not sure how you got from the 3rd line to the 4th line but that is wrong. The modulus of x+iy-i is sqrt{x²+(y-1)²} similarly with the other one

Hence
sqrt{x²+(y-1)²}=sqrt{(x+2)²+(y-3)²}

Then simplify from here

 

[GaDaMIt]

|z-i|=|z+2-3i|


x² + (y-1)² = (x+2)² + (y-3)²

-2y + 1 = 4x + 4 - 6y + 9

4x - 4y = 1 - 4 - 9
= -12
x - y = -3

x - y + 3 = 0


?? right?