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Given complex numbers Z1 = 2 + 3i and Z2 = 2 − i.

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Here's the solution to question 8:

Given complex numbers Z1=2+3iZ_1 = 2 + 3i and Z2=2−iZ_2 = 2 - i.

Part a· 4 steps

Express Z1Z2\frac{Z_1}{Z_2} in the form a+bia + bi where a,b∈Ra, b \in \mathbb{R}.

  1. Substitute the given values of Z1Z_1 and Z2Z_2.

    Z1Z2=2+3i2−i\frac{Z_1}{Z_2} = \frac{2 + 3i}{2 - i}

  2. Multiply the numerator and denominator by the conjugate of the denominator.

    The conjugate of 2−i2 - i is 2+i2 + i.

    2+3i2−i×2+i2+i\frac{2 + 3i}{2 - i} \times \frac{2 + i}{2 + i}

  3. Expand the numerator and the denominator.

    Numerator:

    (2+3i)(2+i)=2(2)+2(i)+3i(2)+3i(i)(2 + 3i)(2 + i) = 2(2) + 2(i) + 3i(2) + 3i(i) =4+2i+6i+3i2= 4 + 2i + 6i + 3i^2

    Since i2=−1i^2 = -1:

    =4+8i−3=1+8i= 4 + 8i - 3 = 1 + 8i

    Denominator:

    (2−i)(2+i)=22−i2(2 - i)(2 + i) = 2^2 - i^2 =4−(−1)=4+1=5= 4 - (-1) = 4 + 1 = 5

  4. Combine the results.

    Z1Z2=1+8i5=15+85i\frac{Z_1}{Z_2} = \frac{1 + 8i}{5} = \frac{1}{5} + \frac{8}{5}i

    The expression in the form a+bia + bi is 15+85i\boxed{\frac{1}{5} + \frac{8}{5}i}.

Part b· 3 steps

Find the modulus and argument of Z1×Z2Z_1 \times Z_2.

  1. Calculate the product Z1×Z2Z_1 \times Z_2.

    Z1×Z2=(2+3i)(2−i)Z_1 \times Z_2 = (2 + 3i)(2 - i) =2(2)+2(−i)+3i(2)+3i(−i)= 2(2) + 2(-i) + 3i(2) + 3i(-i) =4−2i+6i−3i2= 4 - 2i + 6i - 3i^2

    Since i2=−1i^2 = -1:

    =4+4i−3(−1)= 4 + 4i - 3(-1) =4+4i+3=7+4i= 4 + 4i + 3 = 7 + 4i

  2. Find the modulus of Z1×Z2Z_1 \times Z_2.

    For a complex number x+yix + yi, the modulus is x2+y2\sqrt{x^2 + y^2}.

    ∣Z1×Z2∣=∣7+4i∣=72+42|Z_1 \times Z_2| = |7 + 4i| = \sqrt{7^2 + 4^2} =49+16=65= \sqrt{49 + 16} = \sqrt{65}

  3. Find the argument of Z1×Z2Z_1 \times Z_2.

    For a complex number x+yix + yi in the first quadrant, the argument is tan⁡−1(yx)\tan^{-1}\left(\frac{y}{x}\right).
    Since 7+4i7 + 4i is in the first quadrant (real part positive, imaginary part positive):

    arg⁡(Z1×Z2)=tan⁡−1(47)\arg(Z_1 \times Z_2) = \tan^{-1}\left(\frac{4}{7}\right)

    The modulus is 65\boxed{\sqrt{65}} and the argument is tan⁡−1(47)\boxed{\tan^{-1}\left(\frac{4}{7}\right)}.

Part c· 5 steps

Given also that Z3Z_3 is a complex number such that ∣Z3∣=5|Z_3| = 5 and arg⁡(Z3)=π3\arg(Z_3) = \frac{\pi}{3}, express Z1×Z2Z3\frac{Z_1 \times Z_2}{Z_3} in polar form.

  1. From part b), we have Z1×Z2=7+4iZ_1 \times Z_2 = 7 + 4i.

    Its modulus is ∣Z1×Z2∣=65|Z_1 \times Z_2| = \sqrt{65}.
    Its argument is arg⁡(Z1×Z2)=tan⁡−1(47)\arg(Z_1 \times Z_2) = \tan^{-1}\left(\frac{4}{7}\right).

  2. We are given Z3Z_3 in polar form

    ∣Z3∣=5|Z_3| = 5
    arg⁡(Z3)=π3\arg(Z_3) = \frac{\pi}{3}

  3. To find the modulus of the quotient Z1×Z2Z3\frac{Z_1 \times Z_2}{Z_3}, divide their moduli.

    ∣Z1×Z2Z3∣=∣Z1×Z2∣∣Z3∣=655\left|\frac{Z_1 \times Z_2}{Z_3}\right| = \frac{|Z_1 \times Z_2|}{|Z_3|} = \frac{\sqrt{65}}{5}

  4. To find the argument of the quotient Z1×Z2Z3\frac{Z_1 \times Z_2}{Z_3}, subtract their arguments.

    arg⁡(Z1×Z2Z3)=arg⁡(Z1×Z2)−arg⁡(Z3)\arg\left(\frac{Z_1 \times Z_2}{Z_3}\right) = \arg(Z_1 \times Z_2) - \arg(Z_3) =tan⁡−1(47)−π3= \tan^{-1}\left(\frac{4}{7}\right) - \frac{\pi}{3}

  5. Express the result in polar form, r(cos⁡θ+isin⁡θ)r(\cos \theta + i \sin \theta).

    Z1×Z2Z3=655(cos⁡(tan⁡−1(47)−π3)+isin⁡(tan⁡−1(47)−π3))\frac{Z_1 \times Z_2}{Z_3} = \frac{\sqrt{65}}{5} \left( \cos\left(\tan^{-1}\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) + i \sin\left(\tan^{-1}\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) \right)

    The expression in polar form is 655(cos⁡(tan⁡−1(47)−π3)+isin⁡(tan⁡−1(47)−π3))\boxed{\frac{\sqrt{65}}{5} \left( \cos\left(\tan^{-1}\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) + i \sin\left(\tan^{-1}\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) \right)}.

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