JEE Main 2023 (11 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let $w_{1}$ be the point obtained by the rotation of $z_{1} = 5 + 4 i$ about the origin through a right angle in the anticlockwise direction, and $w_{2}$ be the point obtained by the rotation of $z_{2} = 3 + 5 i$ about the origin through a right angle in the clockwise direction. Then the principal argument $w_{1} - w_{2}$ is equal to

Q: Why do we use $\pi - \tan^{-1}(\frac{y}{x})$ for the second quadrant? Can't I just use $\tan^{-1}(\frac{y}{x})$?

That is a classic trap! The function $\tan^{-1}$ only returns values between $-\pi/2$ and $\pi/2$. If you just use $\tan^{-1}(\frac{y}{x})$ for a point like $(-9, 8)$, you will get a negative angle in the fourth quadrant. To 'correct' this and land in the second quadrant, we add $\pi$ to the result, effectively shifting the angle by $180^\circ$ to account for the negative real part.

Q: Is there a way to solve this without complex numbers, maybe using coordinate geometry?

Absolutely! You could rotate the points $(5, 4)$ and $(3, 5)$ using rotation matrices. However, in the JEE Advanced exam, time is your greatest enemy. Complex numbers turn a tedious matrix multiplication into a simple algebraic product. Mastering this 'complex' shortcut is exactly what separates the top rankers from the rest.

Select Answer:

Visualized Solution

Plotting $z_{1}$ and $z_{2}$

Given complex numbers: $z_{1} = 5 + 4 i$ and $z_{2} = 3 + 5 i$ .
Both points lie in the first quadrant of the Argand plane.

Rotation of $z_{1}$

$w_{1}$ is obtained by rotating $z_{1}$ by $9 0^{\circ}$ anticlockwise.
Rotation by $+ 9 0^{\circ}$ is equivalent to multiplying by $e^{iπ /2} = i$ .

Calculating $w_{1}$

$w_{1} = z_{1} \cdot i$
$w_{1} = (5 + 4 i) i = 5 i + 4 i^{2}$
Since $i^{2} = - 1$ , $w_{1} = - 4 + 5 i$ .

Rotation of $z_{2}$

$w_{2}$ is obtained by rotating $z_{2}$ by $9 0^{\circ}$ clockwise.
Rotation by $- 9 0^{\circ}$ is equivalent to multiplying by $e^{- iπ /2} = - i$ .

Calculating $w_{2}$

$w_{2} = z_{2} \cdot (- i)$
$w_{2} = (3 + 5 i) (- i) = - 3 i - 5 i^{2}$
Since $i^{2} = - 1$ , $w_{2} = 5 - 3 i$ .

Setting up $w_{1} - w_{2}$

We need to find the principal argument of $w_{1} - w_{2}$ .
Substitute the values: $w_{1} - w_{2} = (- 4 + 5 i) - (5 - 3 i)$ .

Calculating $w_{1} - w_{2}$

Group real and imaginary parts:
Real part: $- 4 - 5 = - 9$
Imaginary part: $5 i - (- 3 i) = 8 i$
$w_{1} - w_{2} = - 9 + 8 i$

Locating the Resultant Vector

Let $Z = - 9 + 8 i$ .
Real part $x = - 9 < 0$ , Imaginary part $y = 8 > 0$ .
The point $(- 9, 8)$ lies in the Second Quadrant.

Principal Argument Formula

For a complex number $Z = x + i y$ in the second quadrant:
Principal Argument $θ = π - tan^{- 1} \frac{y}{x}$

Evaluating the Argument

Substitute $x = - 9$ and $y = 8$ :
$θ = π - tan^{- 1} \frac{8}{- 9}$
$θ = π - tan^{- 1} \frac{8}{9}$

Final Answer

The principal argument of $w_{1} - w_{2}$ is $π - tan^{- 1} \frac{8}{9}$ .
Matches the first option.

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The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

The Geometry of Rotation

Imagine you are standing at the origin of the Argand plane. You have two vectors,

z_{1} = 5 + 4 i

and

z_{2} = 3 + 5 i

. Both are comfortably resting in the first quadrant.

The secret weapon of the JEE topper is the Rotation Theorem. Rotating a complex number

z

by an angle

θ

is equivalent to multiplying it by

e^{i θ}

When

θ = 9 0^{\circ}

(or

π /2

radians),

e^{iπ /2}

is simply

i

. When

θ = - 9 0^{\circ}

(clockwise),

e^{- iπ /2}

- i

The Calculation

For

w_{1}

, we rotate

z_{1}

anticlockwise by

9 0^{\circ}

. We perform the multiplication:

w_{1} = (5 + 4 i) \cdot i = 5 i + 4 i^{2}

Since

i^{2} = - 1

, we obtain:

w_{1} = - 4 + 5 i

For

w_{2}

, we rotate

z_{2}

clockwise by

9 0^{\circ}

. We perform the multiplication:

w_{2} = (3 + 5 i) \cdot (- i) = - 3 i - 5 i^{2}

Substituting

i^{2} = - 1

, we obtain:

w_{2} = 5 - 3 i

The Final Argument

Now, we find the difference between the two rotated vectors:

w_{1} - w_{2} = (- 4 + 5 i) - (5 - 3 i)

Grouping the real and imaginary parts, we get:

w_{1} - w_{2} = (- 4 - 5) + (5 i + 3 i) = - 9 + 8 i

The resulting vector is

- 9 + 8 i

, which corresponds to the coordinates

(- 9, 8)

. Since the real part is negative and the imaginary part is positive, the vector lies in the second quadrant.

To find the principal argument, we use the geometry of the second quadrant:

θ = π - tan^{- 1} (\frac{y}{x})

Substituting our values, we arrive at the final result:

θ = π - tan^{- 1} (\frac{8}{9})

Common Traps & FAQs

Wait, why is clockwise rotation

- i

? Shouldn't it be positive?

In the Argand plane, the positive direction of rotation is defined as anticlockwise, following the standard mathematical convention for angles. Therefore, a clockwise rotation of

9 0^{\circ}

is equivalent to an angle of

- 9 0^{\circ}

. Since

e^{- iπ /2} = cos (- π /2) + i sin (- π /2) = 0 - i = - i

, we multiply by

- i

to achieve a clockwise shift.

Why do we use

π - tan^{- 1} (\frac{y}{x})

for the second quadrant? Can't I just use

tan^{- 1} (\frac{y}{x})

That is a classic trap! The function

tan^{- 1}

only returns values between

- π /2

and

π /2

. If you just use

tan^{- 1} (\frac{y}{x})

for a point like

(- 9, 8)

, you will get a negative angle in the fourth quadrant. To 'correct' this and land in the second quadrant, we add

π

to the result, effectively shifting the angle by

18 0^{\circ}

to account for the negative real part.

Is there a way to solve this without complex numbers, maybe using coordinate geometry?

Absolutely! You could rotate the points

(5, 4)

and

(3, 5)

using rotation matrices. However, in the JEE Advanced exam, time is your greatest enemy. Complex numbers turn a tedious matrix multiplication into a simple algebraic product. Mastering this 'complex' shortcut is exactly what separates the top rankers from the rest.

Select Answer:

Visualized Solution

Plotting $z_{1}$ and $z_{2}$

Rotation of $z_{1}$

Calculating $w_{1}$

Rotation of $z_{2}$

Calculating $w_{2}$

Setting up $w_{1} - w_{2}$

Calculating $w_{1} - w_{2}$

Locating the Resultant Vector

Principal Argument Formula

Evaluating the Argument

Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Rotation

The Calculation

The Final Argument

Similar Questions

Let a circle $C$ in complex plane pass through the points $z_{1} = 3 + 4 i$ , $z_{2} = 4 + 3 i$ and $z_{3} = 5 i$ . If $z (\neq = z_{1})$ is a point on $C$ such that the line through $z$ and $z_{1}$ is perpendicular to the line through $z_{2}$ and $z_{3}$ , then $ar g (z)$ is equal to :

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 $z_{1}$ 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

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 $0 < t < 1$ .

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let $ar g (z)$ represent the principal argument of the complex number $z$ . The, $∣ z ∣ = 3$ and $ar g (z - 1) - ar g (z + 1) = \frac{π}{4}$ intersect:

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 $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :

Select Answer:

Visualized Solution

Plotting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

Rotation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​

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

Rotation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

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

Setting up .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w1​−w2​

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

Locating the Resultant Vector

Principal Argument Formula

Evaluating the Argument

Final Answer

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Rotation

The Calculation

The Final Argument

Similar Questions

.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; } 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; } 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; } For a non-zero complex number z, let arg(z) denote the principal argument with −π<arg(z)≤π. Then, which of the following statement(s) is (are) FALSE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let arg(z) represent the principal argument of the complex number z. The, ∣z∣=3 and arg(z−1)−arg(z+1)=4π​ intersect:

.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; } Let z1​=10+6i and z2​=4+6i. If z is any complex number such that the argument of z−z2​z−z1​​ is 4π​, then prove that ∣z−7−9i∣=32​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be the complex number satisfying ∣z−5∣≤3 and having maximum positive principal argument. Then 34​5iz+165z−12​​2 is equal to :

Plotting $z_{1}$ and $z_{2}$

Rotation of $z_{1}$

Calculating $w_{1}$

Rotation of $z_{2}$

Calculating $w_{2}$

Setting up $w_{1} - w_{2}$

Calculating $w_{1} - w_{2}$

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 $z_{1}$ 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

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 $0 < t < 1$ .

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let $ar g (z)$ represent the principal argument of the complex number $z$ . The, $∣ z ∣ = 3$ and $ar g (z - 1) - ar g (z + 1) = \frac{π}{4}$ intersect:

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 $z_{1} = 10 + 6 i$ and $z_{2} = 4 + 6 i$ . If $z$ is any complex number such that the argument of $\frac{z - z _{1}}{z - z _{2}}$ is $\frac{π}{4}$ , then prove that $∣ z - 7 - 9 i ∣ = 32$ .

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :