Wait, why is $Re(z_1\bar{z}_2) = 0$ equivalent to saying the vectors are orthogonal?

Great question! In the complex plane, the dot product of two vectors $z_1 = (a, b)$ and $z_2 = (c, d)$ is $ac + bd$. The product $z_1\bar{z}_2$ is $(a + ib)(c - id) = (ac + bd) + i(bc - ad)$. The real part is exactly the dot product. If the real part is zero, the dot product is zero, which is the definition of orthogonality.

Can I just use polar form $z_1 = e^{i\theta_1}$ and $z_2 = e^{i\theta_2}$?

Absolutely! That is a very elegant approach. If $z_1 = e^{i\theta_1}$ and $z_2 = e^{i\theta_2}$, then $z_1\bar{z}_2 = e^{i(\theta_1 - \theta_2)}$. The real part is $\cos(\theta_1 - \theta_2) = 0$, meaning $\theta_1 - \theta_2 = \pm \pi/2$. This confirms they are perpendicular. You can then express $a, b, c, d$ in terms of $\sin$ and $\cos$ and verify the result. Both methods are valid!

JEE Advanced 1985

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: If $z_{1} = a + ib$ and $z_{2} = c + i d$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $R e (z_{1} \overset{z}{ˉ}_{2}) = 0$ , then the pair of complex numbers $w_{1} = a + i c$ and $w_{2} = b + i d$ satisfies

Q: Why do we assume $b^2 = c^2$?

We don't assume it; we derive it! From $a^2 + b^2 = 1$ and $ac + bd = 0$, we found $a = \lambda b$ and $d = -\lambda c$. Substituting these into the unit circle equations gives $b^2(1 + \lambda^2) = 1$ and $c^2(1 + \lambda^2) = 1$. Since both expressions equal 1, they must be equal to each other, forcing $b^2 = c^2$.

Select Answer:

* Multiple Correct

Visualized Solution

Visualizing the Unit Circle

Given $∣ z_{1} ∣ = 1$ and $∣ z_{2} ∣ = 1$ .
This implies $z_{1} = a + ib$ and $z_{2} = c + i d$ lie on the unit circle.
Algebraically: $a^{2} + b^{2} = 1$ and $c^{2} + d^{2} = 1$ .

The Orthogonality Condition

Given $R e (z_{1} \overset{z}{ˉ}_{2}) = 0$ .
$(a + ib) (c - i d) = (a c + b d) + i (b c - a d)$ .
Setting the real part to zero: $a c + b d = 0$ .
This means the vectors $(a, b)$ and $(c, d)$ are orthogonal.

Establishing the Ratio $λ$

From $a c + b d = 0$ , we get $a c = - b d$ .
Rearranging: $\frac{a}{b} = - \frac{d}{c} = λ$ .
Therefore, $a = λb$ and $d = - λ c$ .

Finding the relation between $b$ and $c$

We know $a^{2} + b^{2} = 1$ and $c^{2} + d^{2} = 1$ .
Substitute $a = λb$ : $λ^{2} b^{2} + b^{2} = 1 \Rightarrow b^{2} (1 + λ^{2}) = 1$ .
Substitute $d = - λ c$ : $c^{2} + λ^{2} c^{2} = 1 \Rightarrow c^{2} (1 + λ^{2}) = 1$ .
Comparing the two equations, we get $b^{2} = c^{2}$ .

Evaluating $∣ w_{1} ∣$

Consider the new complex number $w_{1} = a + i c$ .
Its magnitude squared is $∣ w_{1} ∣^{2} = a^{2} + c^{2}$ .
Substitute $a = λb$ : $∣ w_{1} ∣^{2} = λ^{2} b^{2} + c^{2}$ .
Since $b^{2} = c^{2}$ , this becomes $b^{2} (λ^{2} + 1)$ .
From earlier, $b^{2} (λ^{2} + 1) = 1$ , so $∣ w_{1} ∣ = 1$ .

Evaluating $∣ w_{2} ∣$

Consider $w_{2} = b + i d$ .
Its magnitude squared is $∣ w_{2} ∣^{2} = b^{2} + d^{2}$ .
Substitute $d = - λ c$ : $∣ w_{2} ∣^{2} = b^{2} + λ^{2} c^{2}$ .
Since $c^{2} = b^{2}$ , this becomes $b^{2} (1 + λ^{2})$ .
Which equals $1$ , so $∣ w_{2} ∣ = 1$ .

Checking Orthogonality of $w_{1}, w_{2}$

Check the angle between $w_{1}$ and $w_{2}$ by evaluating $R e (w_{1} \overset{w}{ˉ}_{2})$ .
$w_{1} \overset{w}{ˉ}_{2} = (a + i c) (b - i d) = (ab + c d) + i (b c - a d)$ .
So, $R e (w_{1} \overset{w}{ˉ}_{2}) = ab + c d$ .
Substitute $a = λb$ and $d = - λ c$ : $ab + c d = (λb) b + c (- λ c) = λ b^{2} - λ c^{2}$ .
Since $b^{2} = c^{2}$ , this equals $λ (b^{2} - b^{2}) = 0$ .

Conclusion & Takeaways

Key Takeaways:
1. $∣ w_{1} ∣ = 1$ (Option A is correct)
2. $∣ w_{2} ∣ = 1$ (Option B is correct)
3. $R e (w_{1} \overset{w}{ˉ}_{2}) = 0$ (Option C is correct)
This transformation preserves both the unit magnitude and the orthogonality of the pair.

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

We are given two complex numbers,

z_{1} = a + ib

and

z_{2} = c + i d

, both resting on the unit circle. This implies their magnitudes are unity, leading to the constraints:

a^{2} + b^{2} = 1

c^{2} + d^{2} = 1

We are further given the condition

R e (z_{1} \overset{z}{ˉ}_{2}) = 0

. Expanding the product

z_{1} \overset{z}{ˉ}_{2}

yields:

(a + ib) (c - i d) = (a c + b d) + i (b c - a d)

The real part is

a c + b d = 0

. Geometrically, this indicates that the vectors

(a, b)

and

(c, d)

are orthogonal (perpendicular) to each other.

The Power of Substitution

Since

a c + b d = 0

, we can rearrange this to

a c = - b d

. This implies the existence of a common ratio

λ

\frac{a}{b} = - \frac{d}{c} = λ

We can express the components as

a = λb

and

d = - λ c

. Substituting

a = λb

into the first constraint gives:

b^{2} (λ^{2} + 1) = 1

Similarly, substituting

d = - λ c

into the second constraint yields:

c^{2} (1 + λ^{2}) = 1

Because both expressions equal

1

, we arrive at the hidden structural truth:

b^{2} = c^{2}

The Transformation

We evaluate the new complex numbers

w_{1} = a + i c

and

w_{2} = b + i d

. First, we check the magnitude of

w_{1}

∣ w_{1} ∣^{2} = a^{2} + c^{2} = (λb)^{2} + c^{2}

Substituting

c^{2} = b^{2}

, we obtain:

∣ w_{1} ∣^{2} = λ^{2} b^{2} + b^{2} = b^{2} (λ^{2} + 1) = 1

Thus,

∣ w_{1} ∣ = 1

. Applying the same logic to

w_{2}

∣ w_{2} ∣^{2} = b^{2} + d^{2} = b^{2} + (- λ c)^{2} = b^{2} + λ^{2} c^{2}

Substituting

c^{2} = b^{2}

again, we find

∣ w_{2} ∣^{2} = b^{2} (1 + λ^{2}) = 1

. Both

w_{1}

and

w_{2}

lie on the unit circle.

Final Verification of Orthogonality

Finally, we check the orthogonality of

w_{1}

and

w_{2}

by evaluating

R e (w_{1} \overset{w}{ˉ}_{2})

. Expanding

(a + i c) (b - i d)

, the real part is

ab + c d

Substituting

a = λb

and

d = - λ c

R e (w_{1} \overset{w}{ˉ}_{2}) = (λb) b + c (- λ c) = λ b^{2} - λ c^{2}

Since

b^{2} = c^{2}

, the expression simplifies to:

λ (b^{2} - b^{2}) = 0

The orthogonality is preserved. We have demonstrated that this transformation is a symmetry that preserves the fundamental geometric properties of the original pair.

Common Traps & FAQs

Wait, why is

R e (z_{1} \overset{z}{ˉ}_{2}) = 0

equivalent to saying the vectors are orthogonal?

Great question! In the complex plane, the dot product of two vectors

z_{1} = (a, b)

and

z_{2} = (c, d)

a c + b d

. The product

z_{1} \overset{z}{ˉ}_{2}

(a + ib) (c - i d) = (a c + b d) + i (b c - a d)

. The real part is exactly the dot product. If the real part is zero, the dot product is zero, which is the definition of orthogonality.

Can I just use polar form

z_{1} = e^{i θ_{1}}

and

z_{2} = e^{i θ_{2}}

Absolutely! That is a very elegant approach. If

z_{1} = e^{i θ_{1}}

and

z_{2} = e^{i θ_{2}}

, then

z_{1} \overset{z}{ˉ}_{2} = e^{i (θ_{1} - θ_{2})}

. The real part is

cos (θ_{1} - θ_{2}) = 0

, meaning

θ_{1} - θ_{2} = \pm π /2

. This confirms they are perpendicular. You can then express

a, b, c, d

in terms of

sin

and

cos

and verify the result. Both methods are valid!

Why do we assume

b^{2} = c^{2}

We don't assume it; we derive it! From

a^{2} + b^{2} = 1

and

a c + b d = 0

, we found

a = λb

and

d = - λ c

. Substituting these into the unit circle equations gives

b^{2} (1 + λ^{2}) = 1

and

c^{2} (1 + λ^{2}) = 1

. Since both expressions equal 1, they must be equal to each other, forcing

b^{2} = c^{2}

Animated Solution for Mathematics - Complex Numbers: If $z_{1} = a + ib$ and $z_{2} = c + i d$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = 1$ and $R e (z_{1} \overset{z}{ˉ}_{2}) = 0$ , then the pair of complex numbers $w_{1} = a + i c$ and $w_{2} = b + i d$ satisfies

Select Answer:

Visualized Solution

Visualizing the Unit Circle

The Orthogonality Condition

Establishing the Ratio $λ$

Finding the relation between $b$ and $c$

Evaluating $∣ w_{1} ∣$

Evaluating $∣ w_{2} ∣$

Checking Orthogonality of $w_{1}, w_{2}$

Conclusion & Takeaways

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Power of Substitution

The Transformation

Final Verification of Orthogonality

Similar Questions

The number of complex numbers z, satisfying $∣ z ∣ = 1$ and $∣ \frac{z}{z} + \frac{z}{z} ∣ = 1$ is:

Let $z_{1}$ and $z_{2}$ be two distinct complex numbers and let $z = (1 - t) z_{1} + t z_{2}$ for some real number $t$ with $0 < t < 1$ .

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $a r g z_{1} - a r g z_{2}$ is equal to

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq = z_{2}$ and $∣ z_{1} ∣ = ∣ z_{2} ∣$ . If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $(z_{1} + z_{2}) / (z_{1} - z_{2})$ may be

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

If $z_{1}$ and $z_{2}$ are two nonzero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $A r g z_{1} - A r g z_{2}$ is equal to

If $a, b, c$ and $u, v, w$ are complex numbers representing the vertices of two triangles such that $c = (1 - r) a + r b$ and $w = (1 - r) u + r v$ , where $r$ is a complex number, then the two triangles

Let $z$ be a complex number such that $∣ z + 2∣ = 1$ and $Im (\frac{z + 1}{z + 2}) = \frac{1}{5}$ . Then the value of $∣ Re (\overset{z}{ˉ} + 2) ∣$ is

Select Answer:

Visualized Solution

Visualizing the Unit Circle

The Orthogonality Condition

Establishing the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the relation between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣w1​∣

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣w2​∣

Checking Orthogonality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w1​,w2​

Conclusion & Takeaways

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Power of Substitution

The Transformation

Final Verification of Orthogonality

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of complex numbers z, satisfying ∣z∣=1 and ∣zz​+zz​∣=1 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be two distinct complex numbers and let z=(1−t)z1​+tz2​ for some real number t with 0<t<1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of complex numbers z such that ∣z−1∣=∣z+1∣=∣z−i∣ equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​ and z2​ are two non-zero complex numbers such that ∣z1​+z2​∣=∣z1​∣+∣z2​∣, then argz1​−argz2​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are complex numbers such that Re(z1​)=∣z1​−1∣, Re(z2​)=∣z2​−1∣, and arg(z1​−z2​)=6π​, then Im(z1​+z2​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are two distinct complex number such that ​21​−z1​zˉ2​z1​−2z2​​​=2, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​ and z2​ are two nonzero complex numbers such that ∣z1​+z2​∣=∣z1​∣+∣z2​∣, then Argz1​−Argz2​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number such that ∣z+2∣=1 and Im(z+2z+1​)=51​. Then the value of ∣Re(zˉ+2)∣ is

Establishing the Ratio $λ$

Finding the relation between $b$ and $c$

Evaluating $∣ w_{1} ∣$

Evaluating $∣ w_{2} ∣$

Checking Orthogonality of $w_{1}, w_{2}$

The number of complex numbers z, satisfying $∣ z ∣ = 1$ and $∣ \frac{z}{z} + \frac{z}{z} ∣ = 1$ is:

Let $z_{1}$ and $z_{2}$ be two distinct complex numbers and let $z = (1 - t) z_{1} + t z_{2}$ for some real number $t$ with $0 < t < 1$ .

The number of complex numbers $z$ such that $∣ z - 1∣ = ∣ z + 1∣ = ∣ z - i ∣$ equals

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $a r g z_{1} - a r g z_{2}$ is equal to

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

If $z_{1}$ and $z_{2}$ are two nonzero complex numbers such that $∣ z_{1} + z_{2} ∣ = ∣ z_{1} ∣ + ∣ z_{2} ∣$ , then $A r g z_{1} - A r g z_{2}$ is equal to

Let $z$ be a complex number such that $∣ z + 2∣ = 1$ and $Im (\frac{z + 1}{z + 2}) = \frac{1}{5}$ . Then the value of $∣ Re (\overset{z}{ˉ} + 2) ∣$ is