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Hi. I need to find modulus and arguement of the following complex number:

z = -sqrt(2) - sqrt(2i)

I use the following rule to find modulus: \z\ = sqrt(x^2+y^2)

I use the following rule to find argument: arg(z) = tan^-1(y/x)

Please help.

 Jan 28, 2015

Best Answer 

 #1
avatar+100483 
+5

a + bi = -√2 - √2i   (the "i" isn't under the root)

The modulus is given by l z l  = √ (a^2 + b^2) = √[(-√2)^2 + (-√2)^2 ] = √(2 + 2) = √4 = 2

And the arg(z), Θ, is given by

tan-1 (b/a) = tan -1 (-√2/-√2) = tan-1(1) = 5pi/4 + n(2pi)  .... for n = 0,±1,±2, ±3, ±4...

{remember that we're in the 3rd quadrant}

 

 Jan 28, 2015
 #1
avatar+100483 
+5
Best Answer

a + bi = -√2 - √2i   (the "i" isn't under the root)

The modulus is given by l z l  = √ (a^2 + b^2) = √[(-√2)^2 + (-√2)^2 ] = √(2 + 2) = √4 = 2

And the arg(z), Θ, is given by

tan-1 (b/a) = tan -1 (-√2/-√2) = tan-1(1) = 5pi/4 + n(2pi)  .... for n = 0,±1,±2, ±3, ±4...

{remember that we're in the 3rd quadrant}

 

CPhill Jan 28, 2015
 #2
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0

Thank you so much! - you are a lifesaver! 

I have also some other 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

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