Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 months ago

# Write the complex number in polar form with argument θ between 0 and 2π.?

7+8i

Relevance

z=a+ib=7+8i

Polar form

ρ=√(a²+b²)=√113

To identify the angle θ, it should be noted that in [0.2π] there are two tangents that take the value a / b.

The rule states

• IF a>0 THEN θ=arctan(b/a) which is our case.

• IF a<0 THEN θ=π+arctan(b/a)

So il polar form

7 + 8i=√113 (cos(arctan(8/7) + isin(arctan(8/7))

Check it out.

WE have to remember that:

cos(arctan(x))=1/√(1+x²)

sin(arctan(x))=x/√(1+x²)

hence

cos(arctan(8/7))=1/√(1+(8/7)²)=

=1/√(113)/7

sin(arctan(8/7))=x/√(1+(8/7)²)=8/√(113)

so

√113(1/√(113)/7+i8/√(113))= 7 + 8i OK!

• --

r = 7 + 8 i

r ² = [ 7² + 8² ] = 49 + 64

| r | = √113

tan ∅ = 8/7

∅ = 49°

z = √113 /_ 49°

• Recall:

z = a + ib ← this is a complex number

m = √(a² + b²) ← this is its modulus

tan(α) = b/a → then you can deduce α ← this is the argument

z = 7 + 8i ← you can see that: a = 7 and you can see that: b = 8

m = √[(7)² + (8)²]

m = √(49 + 64)

m = √113 ← this is the modulus of z

tan(α) = b/a

tan(α) = 8/7

α ≈ 48.814 ° ≈ 0.852 rd

z = m.[cos(α) + i.sn(α)]

z = (√113).[cos(0.852) + i.sn(0.852)]

z = (√113 ; 0.852) ← polar form

• The polar form of a complex number z=a+bi is z=r(cosθ+isinθ), which we might simplify to r cis(θ).

r = √(7² + 8²) = √113 and θ (in Q1) is arctan(8/7).

7 + 8i = √113 cis(arctan(8/7))

You can find approximate values from here if required.