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

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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)

Jan 28, 2015

#1
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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
+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}

CPhill Jan 28, 2015
#2
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