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Find the argument of 2+3i

Mathematics

ScanSolve AI Answer

Answer

56.31∘ or 0.9828 radians56.31^{\circ} \text{ or } 0.9828 \text{ radians}

Step-by-step solution

3 steps

To find the argument of a complex number z=x+yiz = x + yi, we use the formula arg⁡(z)=arctan⁡(yx)\arg(z) = \arctan\left(\frac{y}{x}\right), considering the quadrant of the complex number.

  1. Identify the real and imaginary parts of the complex number.

    The given complex number is z=2+3iz = 2 + 3i.
    Here, the real part is x=2x = 2 and the imaginary part is y=3y = 3.

  2. Determine the quadrant of the complex number.

    Since x>0x > 0 and y>0y > 0, the complex number 2+3i2+3i lies in the first quadrant. Therefore, the principal argument will be directly given by arctan⁡(yx)\arctan\left(\frac{y}{x}\right).

  3. Calculate the argument.

    arg⁡(z)=arctan⁡(32)\arg(z) = \arctan\left(\frac{3}{2}\right)

    Using a calculator,

    arg⁡(z)≈56.31∘\arg(z) \approx 56.31^{\circ}

    In radians,

    arg⁡(z)≈0.9828 radians\arg(z) \approx 0.9828 \text{ radians}

    The argument of 2+3i2+3i is approximately:

    56.31∘ or 0.9828 radians56.31^{\circ} \text{ or } 0.9828 \text{ radians}
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