A complex number question-How do i express Z= -2 - i in the modulus-argument form?

Answer 1

The modulus argument form is the same as the polar form. There are several different notations you can use.

The magnitude is the same as the modulus.

|Z| = sqrt((-2)^2 + (-1)^2) = sqrt(5) = 2.24

The angle is the same as the argument.

theta = arctan(-1/-2) = -153 degrees = -2.68 radians

You can use an angle symbol to separate the magnitude from the angle.
Z = 2.24 /_ -2.68.

Or you can represent the complex number as an exponential.

Z = 2.24*exp(i*-2.68)

Answer 2

r = √(2^2 +1^2) = √5
∅ = arctan -1/-2 = .46365

Z = r e^(i ∅ + 2n pi i) = √5 e^[(.46465 + pi + 2n pi )i]

Reasons: + pi becaue -2-i is in the third quadrant. + 2n pi because the period of the function is 2 pi.

Answer 3

for : Z = a + bi
modulus(z) ... |z| =√(a2+b2)
and
argument(z) = Tan-1(b/a)

then |z| = √5 and argument(z) = -153.435 deg
we used -153.435 and not 26.565 as a and b are both negative ... (the 3rd quarter)

Answer 4

r=sqrt(4+1)=sqrt(5)
tanx=-1/-2=1/2

z=r(cosx +isinx)
=sqrt(5) . (-cos (arctan(1/2)-i sin(arctan(1/2))

z=(r,x)
=(sqrt(5),arctan(1/2)+pi)