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Solve the equation E in the set of complex numbers.

Mathematics

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Answer

3−14+i3+14\frac{\sqrt{3}-1}{4} + i\frac{\sqrt{3}+1}{4}

22(cos⁡5π12+isin⁡5π12)\frac{\sqrt{2}}{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right)

Step-by-step solution

3 steps

We are given two complex numbers Z1=1+iZ_1 = 1+i and Z2=3−iZ_2 = \sqrt{3}-i. We need to find Z1Z2\frac{Z_1}{Z_2} in both algebraic and polar forms.

Part i) Write Z1Z2\frac{Z_1}{Z_2} in algebraic form and in polar form.

  1. Calculate Z1Z2\frac{Z_1}{Z_2} in algebraic form.

    To express Z1Z2\frac{Z_1}{Z_2} in algebraic form, we multiply the numerator and denominator by the conjugate of the denominator.

    Z1Z2=1+i3−i\frac{Z_1}{Z_2} = \frac{1+i}{\sqrt{3}-i}

    Multiply by 3+i3+i\frac{\sqrt{3}+i}{\sqrt{3}+i}:

    1+i3−i×3+i3+i=(1+i)(3+i)(3−i)(3+i)\frac{1+i}{\sqrt{3}-i} \times \frac{\sqrt{3}+i}{\sqrt{3}+i} = \frac{(1+i)(\sqrt{3}+i)}{(\sqrt{3}-i)(\sqrt{3}+i)}

    Expand the numerator:

    (1+i)(3+i)=1⋅3+1⋅i+i⋅3+i2=3+i+i3−1=(3−1)+i(3+1)(1+i)(\sqrt{3}+i) = 1 \cdot \sqrt{3} + 1 \cdot i + i \cdot \sqrt{3} + i^2 = \sqrt{3} + i + i\sqrt{3} - 1 = (\sqrt{3}-1) + i(\sqrt{3}+1)

    Expand the denominator:

    (3−i)(3+i)=(3)2−i2=3−(−1)=3+1=4(\sqrt{3}-i)(\sqrt{3}+i) = (\sqrt{3})^2 - i^2 = 3 - (-1) = 3+1 = 4

    Combine the numerator and denominator:

    Z1Z2=(3−1)+i(3+1)4=3−14+i3+14\frac{Z_1}{Z_2} = \frac{(\sqrt{3}-1) + i(\sqrt{3}+1)}{4} = \boxed{\frac{\sqrt{3}-1}{4} + i\frac{\sqrt{3}+1}{4}}

  2. Convert Z1Z_1 and Z2Z_2 to polar form.

    For Z1=1+iZ_1 = 1+i:
    The modulus is r1=∣Z1∣=12+12=2r_1 = |Z_1| = \sqrt{1^2 + 1^2} = \sqrt{2}.
    The argument is θ1=arg⁡(Z1)\theta_1 = \arg(Z_1). Since Z1Z_1 is in the first quadrant, tan⁡θ1=11=1\tan \theta_1 = \frac{1}{1} = 1, so θ1=π4\theta_1 = \frac{\pi}{4}.
    Thus, Z1=2(cos⁡π4+isin⁡π4)Z_1 = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right).

    For Z2=3−iZ_2 = \sqrt{3}-i:
    The modulus is r2=∣Z2∣=(3)2+(−1)2=3+1=4=2r_2 = |Z_2| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = \sqrt{4} = 2.
    The argument is θ2=arg⁡(Z2)\theta_2 = \arg(Z_2). Since Z2Z_2 is in the fourth quadrant, tan⁡θ2=−13\tan \theta_2 = \frac{-1}{\sqrt{3}}, so θ2=−π6\theta_2 = -\frac{\pi}{6}.
    Thus, Z2=2(cos⁡(−π6)+isin⁡(−π6))Z_2 = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right).

  3. Calculate Z1Z2\frac{Z_1}{Z_2} in polar form.

    To divide complex numbers in polar form, we divide their moduli and subtract their arguments:

    Z1Z2=r1r2(cos⁡(θ1−θ2)+isin⁡(θ1−θ2))\frac{Z_1}{Z_2} = \frac{r_1}{r_2} \left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right)

    Substitute the moduli:

    r1r2=22\frac{r_1}{r_2} = \frac{\sqrt{2}}{2}

    Substitute and subtract the arguments:

    θ1−θ2=π4−(−π6)=π4+π6=3π12+2π12=5π12\theta_1 - \theta_2 = \frac{\pi}{4} - \left(-\frac{\pi}{6}\right) = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi}{12} + \frac{2\pi}{12} = \frac{5\pi}{12}

    Combine to get the polar form:

    Z1Z2=22(cos⁡5π12+isin⁡5π12)\frac{Z_1}{Z_2} = \boxed{\frac{\sqrt{2}}{2}\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right)}
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