1.

Find the modulus of(2+3i)-(5i-4).can someone please answer this?​

Answer»

Answer:

hope this helps

Step-by-step explanation:

Calculation:

z = (2+3i)-(5i-4)

Result:

Rectangular form:

z = 6-2i

Angle notation (phasor):

z = 6.3245553 ∠ -18°26'6″

Polar form:

z = 6.3245553 × (cos (-18°26'6″) + i sin (-18°26'6″))

Exponential form:

z = 6.3245553 × ei (-0.1024164)

Polar coordinates:

R = |z| = 6.3245553 ... magnitude (modulus, ABSOLUTE VALUE)

θ = arg z = -0.3217506 rad = -18.43495° = -18°26'6″ = -0.1024164π rad ... angle (argument or phase)

Cartesian coordinates:

Cartesian form of imaginary number: z = 6-2i

Real part: x = Re z = 6

Imaginary part: y = Im z = -2

Calculation steps

Complex number: 2+3i

Subtract: 5i - 4 = -4+5i

Subtract: the result of step No. 1 - the result of step No. 2 = (2+3i) - (-4+5i) = (2+4) + (3-5)i = 6-2i


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