Question

If $z_1 = a_1 + ib_1$ and $z_2 = a_2 + ib_2$ are complex numbers such that $|z_1| = 1, |z_2| = 2$ and $Re(z_1\overline{z_2}) = 0$, then the pair of complex numbers $w_1 = a_1 + \frac{ia_2}{2}$ and $w_2 = 2b_1 + ib_2$ satisfy. Then

• A. $|w_1| = 1$
• B. $|w_2| = 2$
• C. $Re(w_1w_2) = 0$
• D. $|Im(w_1w_2)| = 2$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

Let the given complex numbers be $z_1 = a_1 + ib_1$ and $z_2 = a_2 + ib_2$. We are given the following conditions:

1. $|z_1| = 1 \implies \sqrt{a_1^2 + b_1^2} = 1 \implies a_1^2 + b_1^2 = 1$ (Equation 1)
2. $|z_2| = 2 \implies \sqrt{a_2^2 + b_2^2} = 2 \implies a_2^2 + b_2^2 = 4$ (Equation 2)
3. $Re(z_1\overline{z_2}) = 0$

First, calculate $z_1\overline{z_2}$: $z_1\overline{z_2} = (a_1 + ib_1)(a_2 - ib_2) = a_1a_2 - ia_1b_2 + ib_1a_2 - i^2b_1b_2$ $z_1\overline{z_2} = (a_1a_2 + b_1b_2) + i(b_1a_2 - a_1b_2)$

Since $Re(z_1\overline{z_2}) = 0$, we have: $a_1a_2 + b_1b_2 = 0$ (Equation 3)

From Equation 3, we can derive relations between $a_1, b_1, a_2, b_2$. If $a_2 \ne 0$, then $a_1 = -\frac{b_1b_2}{a_2}$. Substitute this into Equation 1: $(-\frac{b_1b_2}{a_2})^2 + b_1^2 = 1 \implies \frac{b_1^2b_2^2}{a_2^2} + b_1^2 = 1 \implies b_1^2(\frac{b_2^2}{a_2^2} + 1) = 1$ $b_1^2(\frac{b_2^2 + a_2^2}{a_2^2}) = 1$. Using Equation 2 ($a_2^2 + b_2^2 = 4$): $b_1^2(\frac{4}{a_2^2}) = 1 \implies 4b_1^2 = a_2^2$. This relation also holds if $a_2=0$ (which implies $b_1=0$). So, $a_2^2 = 4b_1^2$ (Relation A)

Similarly, if $b_2 \ne 0$, then $b_1 = -\frac{a_1a_2}{b_2}$. Substitute this into Equation 1: $a_1^2 + (-\frac{a_1a_2}{b_2})^2 = 1 \implies a_1^2 + \frac{a_1^2a_2^2}{b_2^2} = 1 \implies a_1^2(1 + \frac{a_2^2}{b_2^2}) = 1$ $a_1^2(\frac{b_2^2 + a_2^2}{b_2^2}) = 1$. Using Equation 2 ($a_2^2 + b_2^2 = 4$): $a_1^2(\frac{4}{b_2^2}) = 1 \implies 4a_1^2 = b_2^2$. This relation also holds if $b_2=0$ (which implies $a_1=0$). So, $b_2^2 = 4a_1^2$ (Relation B)

Now let's evaluate the options for $w_1 = a_1 + \frac{ia_2}{2}$ and $w_2 = 2b_1 + ib_2$.

A. $|w_1| = 1$ $|w_1|^2 = |a_1 + \frac{ia_2}{2}|^2 = a_1^2 + (\frac{a_2}{2})^2 = a_1^2 + \frac{a_2^2}{4}$. Using Relation A ($a_2^2 = 4b_1^2$): $|w_1|^2 = a_1^2 + \frac{4b_1^2}{4} = a_1^2 + b_1^2$. Using Equation 1 ($a_1^2 + b_1^2 = 1$): $|w_1|^2 = 1 \implies |w_1| = 1$. So, Option A is true.

B. $|w_2| = 2$ $|w_2|^2 = |2b_1 + ib_2|^2 = (2b_1)^2 + b_2^2 = 4b_1^2 + b_2^2$. Using Relation A ($4b_1^2 = a_2^2$) and Relation B ($b_2^2 = 4a_1^2$): $|w_2|^2 = a_2^2 + 4a_1^2$. We also know from Equation 2 that $a_2^2 + b_2^2 = 4$. And from Equation 1 that $a_1^2 + b_1^2 = 1$. Substitute Relation A and Relation B into Equation 2: $4b_1^2 + 4a_1^2 = 4 \implies 4(b_1^2 + a_1^2) = 4 \implies a_1^2 + b_1^2 = 1$, which is consistent with Equation 1. So, $|w_2|^2 = 4b_1^2 + b_2^2 = 4b_1^2 + 4a_1^2 = 4(b_1^2 + a_1^2) = 4(1) = 4$. $|w_2|^2 = 4 \implies |w_2| = 2$. So, Option B is true.

C. $Re(w_1w_2) = 0$ $w_1w_2 = (a_1 + \frac{ia_2}{2})(2b_1 + ib_2)$ $w_1w_2 = 2a_1b_1 + ia_1b_2 + i a_2b_1 - \frac{a_2b_2}{2}$ $Re(w_1w_2) = 2a_1b_1 - \frac{a_2b_2}{2}$. We need to check if $2a_1b_1 - \frac{a_2b_2}{2} = 0 \iff 4a_1b_1 = a_2b_2$. From Relation A, $a_2 = \pm 2b_1$. From Relation B, $b_2 = \pm 2a_1$. Substitute these into Equation 3: $a_1a_2 + b_1b_2 = 0$. $a_1(\pm 2b_1) + b_1(\pm 2a_1) = 0$. If $a_2 = 2b_1$ and $b_2 = 2a_1$: $a_1(2b_1) + b_1(2a_1) = 0 \implies 4a_1b_1 = 0$. If $a_2 = -2b_1$ and $b_2 = -2a_1$: $a_1(-2b_1) + b_1(-2a_1) = 0 \implies -4a_1b_1 = 0$. In both these cases, $a_1b_1 = 0$, which implies $a_1=0$ or $b_1=0$. If $a_1=0$, then $b_2=0$ (from Relation B). If $b_1=0$, then $a_2=0$ (from Relation A). In these scenarios, $a_1b_1=0$ and $a_2b_2=0$, so $4a_1b_1 = a_2b_2 = 0$ holds. Thus $Re(w_1w_2)=0$.

Consider the other two cases for signs: If $a_2 = 2b_1$ and $b_2 = -2a_1$: $a_1(2b_1) + b_1(-2a_1) = 0 \implies 2a_1b_1 - 2a_1b_1 = 0 \implies 0 = 0$. This is always true. In this case, we need to check $4a_1b_1 = a_2b_2$. $4a_1b_1 = (2b_1)(-2a_1) = -4a_1b_1$. This implies $8a_1b_1 = 0$, so $a_1b_1=0$. This means either $a_1=0$ or $b_1=0$. If $a_1=0$, then $b_2=0$ (from $b_2=-2a_1$). From Equation 1, $b_1=\pm 1$. From Equation 2, $a_2=\pm 2$. The condition $a_2=2b_1$ means $a_2$ and $b_1$ have the same sign. If $b_1=1, a_2=2$. Then $a_1=0, b_1=1, a_2=2, b_2=0$. $Re(w_1w_2) = 2(0)(1) - \frac{2(0)}{2} = 0$. If $b_1=-1, a_2=-2$. Then $a_1=0, b_1=-1, a_2=-2, b_2=0$. $Re(w_1w_2) = 2(0)(-1) - \frac{(-2)(0)}{2} = 0$. So $Re(w_1w_2)=0$ holds.

If $a_2 = -2b_1$ and $b_2 = 2a_1$: $a_1(-2b_1) + b_1(2a_1) = 0 \implies -2a_1b_1 + 2a_1b_1 = 0 \implies 0 = 0$. This is always true. In this case, we need to check $4a_1b_1 = a_2b_2$. $4a_1b_1 = (-2b_1)(2a_1) = -4a_1b_1$. This implies $8a_1b_1 = 0$, so $a_1b_1=0$. Again, this means either $a_1=0$ or $b_1=0$. If $a_1=0$, then $b_2=0$ (from $b_2=2a_1$). From Equation 1, $b_1=\pm 1$. From Equation 2, $a_2=\pm 2$. The condition $a_2=-2b_1$ means $a_2$ and $b_1$ have opposite signs. If $b_1=1, a_2=-2$. Then $a_1=0, b_1=1, a_2=-2, b_2=0$. $Re(w_1w_2) = 2(0)(1) - \frac{(-2)(0)}{2} = 0$. If $b_1=-1, a_2=2$. Then $a_1=0, b_1=-1, a_2=2, b_2=0$. $Re(w_1w_2) = 2(0)(-1) - \frac{2(0)}{2} = 0$. So $Re(w_1w_2)=0$ holds.

In all cases, $Re(w_1w_2) = 0$. So, Option C is true.

D. $|Im(w_1w_2)| = 2$ From the calculation for Option C: $w_1w_2 = 2a_1b_1 - \frac{a_2b_2}{2} + i(a_1b_2 + a_2b_1)$. Since $Re(w_1w_2)=0$, the expression simplifies to $w_1w_2 = i(a_1b_2 + a_2b_1)$. So, $Im(w_1w_2) = a_1b_2 + a_2b_1$. We need to check if $|a_1b_2 + a_2b_1| = 2$. Let's use the relations $a_2 = \pm 2b_1$ and $b_2 = \pm 2a_1$. We have four cases for the signs:

1. $a_2 = 2b_1$ and $b_2 = 2a_1$: $Im(w_1w_2) = a_1(2a_1) + (2b_1)b_1 = 2a_1^2 + 2b_1^2 = 2(a_1^2 + b_1^2)$. Using Equation 1 ($a_1^2 + b_1^2 = 1$): $Im(w_1w_2) = 2(1) = 2$. $|Im(w_1w_2)| = |2| = 2$.

2. $a_2 = -2b_1$ and $b_2 = -2a_1$: $Im(w_1w_2) = a_1(-2a_1) + (-2b_1)b_1 = -2a_1^2 - 2b_1^2 = -2(a_1^2 + b_1^2)$. Using Equation 1 ($a_1^2 + b_1^2 = 1$): $Im(w_1w_2) = -2(1) = -2$. $|Im(w_1w_2)| = |-2| = 2$.

3. $a_2 = 2b_1$ and $b_2 = -2a_1$: $Im(w_1w_2) = a_1(-2a_1) + (2b_1)b_1 = -2a_1^2 + 2b_1^2 = 2(b_1^2 - a_1^2)$. This value is not necessarily 2. For example, if $a_1=0, b_1=1$, then $Im(w_1w_2)=2(1-0)=2$. But if $a_1=1, b_1=0$, then $Im(w_1w_2)=2(0-1)=-2$. If $a_1=1/\sqrt{2}, b_1=1/\sqrt{2}$, then $Im(w_1w_2)=2(1/2-1/2)=0$. Since $Im(w_1w_2)$ can be 0, $|Im(w_1w_2)|$ is not always 2. For instance, if $z_1 = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$, $z_2 = 0 + i(2)$. Then $a_1=1/\sqrt{2}, b_1=1/\sqrt{2}, a_2=0, b_2=2$. $|z_1|=1, |z_2|=2$. $Re(z_1\overline{z_2}) = Re((\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})(-2i)) = Re(\frac{-2i}{\sqrt{2}} + \frac{2}{\sqrt{2}}) = \frac{2}{\sqrt{2}} \ne 0$. So this example does not satisfy all conditions.

Let's re-examine the consistency of the relations $a_2 = 2b_1$ and $b_2 = -2a_1$ with $a_1a_2 + b_1b_2 = 0$. $a_1(2b_1) + b_1(-2a_1) = 2a_1b_1 - 2a_1b_1 = 0$. This is always consistent. Now, $Im(w_1w_2) = a_1b_2 + a_2b_1 = a_1(-2a_1) + (2b_1)b_1 = -2a_1^2 + 2b_1^2 = 2(b_1^2 - a_1^2)$. We know $a_1^2+b_1^2=1$. So $b_1^2 = 1-a_1^2$. $Im(w_1w_2) = 2((1-a_1^2) - a_1^2) = 2(1 - 2a_1^2)$. The value of $a_1^2$ can range from 0 to 1. If $a_1^2 = 0$, $Im(w_1w_2) = 2(1) = 2$. If $a_1^2 = 1$, $Im(w_1w_2) = 2(1-2) = -2$. If $a_1^2 = 1/2$, $Im(w_1w_2) = 2(1-1) = 0$. Therefore, $|Im(w_1w_2)|$ is not always 2 in this case. It can be 0. For example, let $a_1=1/\sqrt{2}, b_1=1/\sqrt{2}$. Then $z_1 = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$. Then $a_2=2b_1=\sqrt{2}$, $b_2=-2a_1=-\sqrt{2}$. $z_2 = \sqrt{2} - i\sqrt{2}$. Check conditions: $|z_1|^2 = (1/\sqrt{2})^2 + (1/\sqrt{2})^2 = 1/2+1/2=1$. So $|z_1|=1$. $|z_2|^2 = (\sqrt{2})^2 + (-\sqrt{2})^2 = 2+2=4$. So $|z_2|=2$. $Re(z_1\overline{z_2}) = Re((\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})(\sqrt{2} + i\sqrt{2})) = Re(\frac{1}{\sqrt{2}}(\sqrt{2}+i\sqrt{2}) + \frac{i}{\sqrt{2}}(\sqrt{2}+i\sqrt{2}))$ $= Re(1+i+i-1) = Re(2i) = 0$. All conditions are satisfied. Now for $w_1$ and $w_2$: $w_1 = a_1 + \frac{ia_2}{2} = \frac{1}{\sqrt{2}} + i\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$. $w_2 = 2b_1 + ib_2 = 2\frac{1}{\sqrt{2}} + i(-\sqrt{2}) = \sqrt{2} - i\sqrt{2}$. $w_1w_2 = (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})(\sqrt{2} - i\sqrt{2}) = (\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}})\sqrt{2}(1-i) = (1+i)(1-i) = 1-i^2 = 1+1=2$. Here, $Im(w_1w_2) = 0$. So $|Im(w_1w_2)| = 0 \ne 2$. Thus, Option D is not true.

The options that are true are A, B, and C.