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3-4ihow to solve in complex number trig ?

User MWinstead
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In order to write a complex number in its trig form, you take into account the following transformations:

for a complex number z = a + ib

r = (a² + b²)

θ = tan⁻¹ (b/a)

in trig form the complex number is:

z = r[cosθ + i sin]

Thus, for the given complex number z = 3 - 4i, you have:

r = √((3)² + (-4)²) = √(25) = 5

θ = tan⁻¹ (-4/3) = -0.0927

However, you write the previous angle for an interval in between 0 ≤ θ < 2π. -0.0927 angle is the same as the angle 2π - 0.0927 = 5.356 ≈ 5.36

Finally, the trig form of the complex number is:

z = 5[cos5.36 + i sin5.36]

for a complex number z = -2 + 2i, you obtain:

r = √((-2)² + (2)²) = √(4 + 4) = 2√2

θ = tan⁻¹ (2/(-2)) = -0.78

the previous angle is the same as 2π - 0.78 = 5.49 ≈ 5.50

Hence, the tri form of the given complex number is:

z = 2√2 [cos5.50 + i sin5.50]

For the complex number z = 5 + 3i you have:

r = √((5)²+(3)²) = √34

θ = tan⁻¹(3/5) = 0.54

In this case you don't chane the angle because it is already in between 0 and 2π

Hence, you obtain:

z = √34[cos0.54 + i sin0.54]

User Miljen Mikic
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