Algebra · real student question

Let Z1 and Z2 be complex numbers with conjugate(Z1) times the inverse of Z2 equal to 2, and (1 - i)(conjugate(Z1) + 3) = 4 Z2. Find the modulus of (Z2 times Z1 cubed) divided by (-Z2).

Question

Let Z1Z_1 and Z2Z_2 be complex numbers with Z1‾⋅Z2−1=2\overline{Z_1}\cdot Z_2^{-1}=2 and (1−i)(Z1‾+3)=4Z2(1-i)\left(\overline{Z_1}+3\right)=4Z_2. Find ∣Z2⋅Z13−Z2∣.\left|\frac{Z_2\cdot Z_1^{3}}{-Z_2}\right|.

Step-by-step solution

  1. Simplify the target first. Since Z2≠0Z_2\ne 0 (it has an inverse), the Z2Z_2 factors cancel: Z2Z13−Z2=−Z13,∣−Z13∣=∣−1∣ ∣Z1∣3=∣Z1∣3.\frac{Z_2Z_1^{3}}{-Z_2}=-Z_1^{3},\qquad\left|-Z_1^{3}\right|=|-1|\,|Z_1|^{3}=|Z_1|^{3}. So only the modulus of Z1Z_1 is needed, and Z2Z_2 never has to be found in polar form.

  2. Use the first equation to eliminate Z2Z_2. Z1‾Z2=2\dfrac{\overline{Z_1}}{Z_2}=2 gives Z1‾=2Z2\overline{Z_1}=2Z_2, that is Z2=Z1‾2Z_2=\dfrac{\overline{Z_1}}{2}.

  3. Substitute into the second equation. Writing w=Z1‾w=\overline{Z_1}, (1−i)(w+3)=4⋅w2=2w ⟹ (1−i)w+3−3i=2w.(1-i)(w+3)=4\cdot\frac{w}{2}=2w\ \Longrightarrow\ (1-i)w+3-3i=2w.

  4. Solve the linear equation for ww. Collecting ww: 3−3i=(2−(1−i))w=(1+i)w3-3i=\bigl(2-(1-i)\bigr)w=(1+i)w, so w=3−3i1+i=3(1−i)2(1+i)(1−i)=3(−2i)2=−3i.w=\frac{3-3i}{1+i}=\frac{3(1-i)^2}{(1+i)(1-i)}=\frac{3(-2i)}{2}=-3i.

  5. Recover Z1Z_1 and check the data. Z1‾=−3i\overline{Z_1}=-3i gives Z1=3iZ_1=3i and Z2=−3i2Z_2=-\tfrac{3i}{2}. Check: Z1‾/Z2=(−3i)/(−1.5i)=2\overline{Z_1}/Z_2=(-3i)/(-1.5i)=2, and (1−i)(−3i+3)=3(1−i)2=−6i=4(−1.5i)(1-i)(-3i+3)=3(1-i)^2=-6i=4(-1.5i). Both equations hold.

  6. Take the modulus. ∣Z1∣=∣3i∣=3|Z_1|=|3i|=3, so the answer is ∣Z1∣3=33=27|Z_1|^{3}=3^{3}=27.

Answer

2727

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