how do you write 3-3i in trigonometric form?

Question

could someone go through the steps of writing 3-3i in trig form?

Answer 1

Let Z = 3 - 3i
|Z| = √(3² + 3²)
= 3√2
Thus Z = 3√2(1/√2 - 1/√2i)
Now swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2
Vlearly this is a 4th quadrabt angle
Therefore θ = 2π - π/4
= 7π/4
Thus Z = 3√2(cos7π/4 + i sin7π/4)
[= 3√2e^(i7π/4)]

Answer 2

z = 3-3i

ok trig form is just like a cartesian graph exept with a imaginary axis

First u need 2 find the magnitude
/z/ = (3^2+.3^2)^.5 = 18^.5

Then u need 2 find the argument i.e angle between the line and the x axis

Arg z = tan^-1 (-3/3)
= 45 degrees

.'. z = (18^.5) cis (pi/2)

Answer 3

general trigonometric form of a complex number is
z=r * exp(i * phi)

where r is the modulus of the number

and of course the complex exponential exp (i*phi) is equal to cos(phi)+i*sin(phi).

anyway, so you need to find out the modulus (or absolute value), and the angle, phi.

the modulus you find using Pythagorean theorem you know that you have r=sqrt(a^2+b^2), here your number is z=3-3i so a=3 and b=-3, so this gives you r=sqrt(3^2+3^2)=sqrt(18)=4.243.

the angle you could find by applying some basic trigonometry rule. But clearly you don't even need to do this because with a=b then the angle is clearly simply 45 degrees i.e. pi/4

so you can rewrite your number as

z = sqrt(18)*exp(i*pi/4)


some people, who don't know about the complex exponential function, tend to write "cis(phi)" instead, but I don't see the point, and it blinds them to the properties of the exponential, which can prove too bad in some situations


hope this helps

Answer 4

Trig Form

Answer 5

In general:
Z=a+ib=IZI*[cos(θ)+i*sin(θ)] &
IZI=(a^2+b^2)^½
tan(θ)=b/a

Now it is a straightforward task to rewrite you complex number: 3-i3

IZI = √(3²+3²) = √18 = 3*√2
tan θ = -3/3 = -1 =>
θ= -π/4+p*2*π; p belongs to Z ***

***
When tan has a negative sign § can only lies in 2. and 4. quadrant, and from 3-i3 we can easily see the § lies in 4. quadrant


Summary
3-i3=lZl [θ] =3*√2 [θ = -π/4]
=====================

Answer 6

3 - 3i= sqrt(9 + 9)@arctan(-3/3) = 4.24264@ -π/4

Answer 7

hi, Your expression " Z1=-3+3? 3i " is quite ambiguous. Z? = -3 + 3i?3     = 6 × [-½ + i(?3)/2]     = 6[cos(2?/3) + i.sin(2?/3)]     = 6.e^(2i?/3) Z? = -5i     = 5(0 - i)     = 5[cos(-?/2) + i.sin(-?/2)]     = 5.e^(-?/2) Logically, Dragon.Jade :-)

Answer 8

r = sqrt(3^2+(-3)^2) = 3sqrt(2)

theta = atan (-3/3) = 3pi/4

Thus, it is 3sqrt(2)*exp(3pi*i/4) = 3sqrt(2)(cos 3pi/4 +i*sin 3pi/4)