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I have some complex numbers I need to find modulus and argument for.

  • (1+√3i)(1-√3i)
    • z = a+bi = 1-√3i+√3i+3 = 4
    • lzl  = √(a^2 + b^2) = √4^2 = √16 = 4
    • arg(z) = tan-1(b/a) = tan-1(0/4) = 0 degrees
  • conjugate -i
    • conjugate -i = i
    • lzl  = √(a^2 + b^2) = √1^2 = √1 = 1
    • arg(z) =  tan-1(b/a) = tan-1(1/0) = 90 degrees
  • 1-i/1-i
    • z = a+bi = 1-i/1-i = (1-i)*(1+i)/(1-i)*(1+i) = 1+i-i-i^2/1+i-i-i^2 = 1+i-i+1/1+i-i+1 = 2/2 = 1
    • lzl  = √(a^2 + b^2) = √1^2  = √1 = 1
    • arg(z) =  tan-1(b/a) = tan-1(1/0) = 90 degrees

Is the calculations I did correct? - I think that the last one is wrong, but I just cant figure out why ..

 Jan 28, 2015

Best Answer 

 #1
avatar+98196 
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I think everything is good except the last one...

Note that, similar to the first one, you have z =1 = (1 + 0i)

So, arg(z) = tan-1(0/1) = 0 = 0 degrees (±n*360) for n = 0,1,2,3,4..... 

 

 Jan 28, 2015
 #1
avatar+98196 
+8
Best Answer

I think everything is good except the last one...

Note that, similar to the first one, you have z =1 = (1 + 0i)

So, arg(z) = tan-1(0/1) = 0 = 0 degrees (±n*360) for n = 0,1,2,3,4..... 

 

CPhill Jan 28, 2015

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