Why can't we just use polar form $z = re^{i\theta}$?

You absolutely can! It is actually a very elegant way to solve it. If $z = re^{i\theta}$, then $z^2 = r^2 e^{i2\theta}$ and $i|z|^2 = r^2 e^{i\pi/2}$. Equating them gives $2\theta = \pi/2 + 2n\pi$. This leads to $\theta = \pi/4$ or $5\pi/4$, which correspond to the line $y=x$. Both methods lead to the same beautiful truth.

JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let $z = x + i y$ be a non-zero complex number such that $z^{2} = i ∣ z ∣^{2}$ , where $i = - 1$ , then $z$ lies on the:

Q: Is $z=0$ really excluded?

Yes, the problem explicitly states that $z$ is a non-zero complex number. We must always respect the constraints given in the question. If we included $z=0$, the origin would be a solution, but the problem forces us to look at the lines passing through the origin, excluding the origin itself.

Select Answer:

Visualized Solution

Visualizing the Argand Plane

Let the complex number be $z = x + i y$ where $x, y \in R$ .
Given condition: $z^{2} = i ∣ z ∣^{2}$ and $z \neq = 0$ .

Defining the Components of $z$

$z = x + i y$
$∣ z ∣^{2} = x^{2} + y^{2}$

Setting up the Equation

Substitute $z$ and $∣ z ∣^{2}$ into $z^{2} = i ∣ z ∣^{2}$ :
$(x + i y)^{2} = i (x^{2} + y^{2})$

Expanding the Left Hand Side

Expanding $(x + i y)^{2}$ using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$(x + i y)^{2} = x^{2} + (i y)^{2} + 2 i (x) (y)$

Simplifying the $i^{2}$ Term

Since $i^{2} = - 1$ , then $(i y)^{2} = - y^{2}$
LHS becomes: $x^{2} - y^{2} + 2 i x y$

Grouping Real and Imaginary Parts

Equating the expanded LHS to the RHS:
$x^{2} - y^{2} + i (2 x y) = i (x^{2} + y^{2})$

Equating the Real Parts

For equality, $R e (L H S) = R e (R H S)$
$x^{2} - y^{2} = 0$

Solving the Real Part Equation

$(x - y) (x + y) = 0$
This implies $y = x$ or $y = - x$

Equating the Imaginary Parts

For equality, $I m (L H S) = I m (R H S)$
$2 x y = x^{2} + y^{2}$

Testing Case 1: $y = x$

If $y = x$ , substitute into $2 x y = x^{2} + y^{2}$ :
$2 x (x) = x^{2} + x^{2} \Rightarrow 2 x^{2} = 2 x^{2}$
This is true for all $x \in R$ .

Testing Case 2: $y = - x$

If $y = - x$ , substitute into $2 x y = x^{2} + y^{2}$ :
$2 x (- x) = x^{2} + (- x)^{2} \Rightarrow - 2 x^{2} = 2 x^{2}$

Analyzing the Conflict in Case 2

$- 2 x^{2} = 2 x^{2} \Rightarrow 4 x^{2} = 0$
This implies $x = 0$ , which means $y = 0$ .

Applying the Non-Zero Constraint

Constraint: $z$ is a non-zero complex number ( $z \neq = 0$ ).
Therefore, the solution $x = 0, y = 0$ is excluded.

Final Conclusion

The only valid condition for non-zero $z$ is $y = x$ .
Thus, $z$ lies on the line $y = x$ .

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We are given a non-zero complex number

z = x + i y

that satisfies the condition:

z^{2} = i ∣ z ∣^{2}

Here,

x

and

y

are real numbers representing the coordinates of

z

in the Argand plane. The modulus squared is defined as

∣ z ∣^{2} = x^{2} + y^{2}

The Expansion

Substituting

z = x + i y

into the given equation, we have:

(x + i y)^{2} = i (x^{2} + y^{2})

Expanding the left-hand side using the identity

(a + b)^{2} = a^{2} + 2 ab + b^{2}

, we obtain:

x^{2} + (i y)^{2} + 2 i (x y) = i (x^{2} + y^{2})

Since

i^{2} = - 1

, the expression simplifies to:

x^{2} - y^{2} + i (2 x y) = i (x^{2} + y^{2})

The Great Separation

For two complex numbers to be equal, their real and imaginary parts must be identical. Equating the real parts, we get:

x^{2} - y^{2} = 0

This factors into

(x - y) (x + y) = 0

, which implies that either

y = x

y = - x

. These represent two lines passing through the origin.

The Final Verdict

Next, we equate the imaginary parts:

2 x y = x^{2} + y^{2}

We test our two candidate lines:

1. If

y = x

, the equation becomes

2 x^{2} = x^{2} + x^{2}

, which simplifies to

2 x^{2} = 2 x^{2}

. This is a universal identity that holds for all $x eq 0$.

2. If

y = - x

, the equation becomes

2 x (- x) = x^{2} + (- x)^{2}

, which simplifies to

- 2 x^{2} = 2 x^{2}

, or

4 x^{2} = 0

. This forces

x = 0

and

y = 0

Since the problem specifies that

z

is a non-zero complex number, the origin

(0, 0)

is excluded. Therefore, the only valid solution is the line

y = x

The complex number

z

must lie on the line $y = x$ .

Common Traps & FAQs

Wait, why do we reject the negative root here?

It is not about rejecting a root, but about checking the constraint. The equation

2 x y = x^{2} + y^{2}

must hold true. When we plug in

y = - x

, we get

- 2 x^{2} = 2 x^{2}

, which only works if

x = 0

. Since

z \neq = 0

, that case is invalid.

Why can't we just use polar form

z = r e^{i θ}

You absolutely can! It is actually a very elegant way to solve it. If

z = r e^{i θ}

, then

z^{2} = r^{2} e^{i 2 θ}

and

i ∣ z ∣^{2} = r^{2} e^{iπ /2}

. Equating them gives

2 θ = π /2 + 2 nπ

. This leads to

θ = π /4

5 π /4

, which correspond to the line

y = x

. Both methods lead to the same beautiful truth.

z = 0

really excluded?

Yes, the problem explicitly states that

z

is a non-zero complex number. We must always respect the constraints given in the question. If we included

z = 0

, the origin would be a solution, but the problem forces us to look at the lines passing through the origin, excluding the origin itself.

Animated Solution for Mathematics - Complex Numbers: Let $z = x + i y$ be a non-zero complex number such that $z^{2} = i ∣ z ∣^{2}$ , where $i = - 1$ , then $z$ lies on the:

Select Answer:

Visualized Solution

Visualizing the Argand Plane

Defining the Components of $z$

Setting up the Equation

Expanding the Left Hand Side

Simplifying the $i^{2}$ Term

Grouping Real and Imaginary Parts

Equating the Real Parts

Solving the Real Part Equation

Equating the Imaginary Parts

Testing Case 1: $y = x$

Testing Case 2: $y = - x$

Analyzing the Conflict in Case 2

Applying the Non-Zero Constraint

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Expansion

The Great Separation

The Final Verdict

Similar Questions

$z \neq = 0$ is a complex number

List-I

List-II

Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq = 0$ . If the complex number $z = x + i y$ satisfies $Im (\frac{a z + b}{z + 1}) = y$ , then which of the following is(are) possible value(s) of $x$ ?

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number}$ . Then which of the following is NOT correct?

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2 + 3 z + 4 z ^{2}}{2 - 3 z + 4 z ^{2}}$ is a real number, then the value of $∣ z ∣^{2}$ is _____________.

If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Let $z$ be a complex number such that $∣ z ∣ + z = 3 + i$ (where $i = - 1$ ). Then $∣ z ∣$ is equal to:

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

Let $S$ be the set of all complex numbers $z$ satisfying $∣ z^{2} + z + 1∣ = 1$ . Then which of the following statements is/are TRUE?

Let $\overset{z}{ˉ}$ denote the complex conjugate of a complex number $z$ and let $i = - 1$ . In the set of complex numbers, the number of distinct roots of the equation $\overset{z}{ˉ} - z^{2} = i (\overset{z}{ˉ} + z^{2})$ is _____________.

If for $z = α + i β$ , $∣ z + 2∣ = z + 4 (1 + i)$ , then $α + β$ and $α β$ are the roots of the equation

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z=x+iy be a non-zero complex number such that z2=i∣z∣2, where i=−1​, then z lies on the:

Select Answer:

Visualized Solution

Visualizing the Argand Plane

Defining the Components of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Setting up the Equation

Expanding the Left Hand Side

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i2 Term

Grouping Real and Imaginary Parts

Equating the Real Parts

Solving the Real Part Equation

Equating the Imaginary Parts

Testing Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=x

Testing Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=−x

Analyzing the Conflict in Case 2

Applying the Non-Zero Constraint

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Expansion

The Great Separation

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=0 is a complex number

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z=x+iy:4z+2i2z−3i​ is a real number}. Then which of the following is NOT correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number with non-zero imaginary part. If 2−3z+4z22+3z+4z2​ is a real number, then the value of ∣z∣2 is _____________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=x−iy and z1/3=p+iq, then (x/p+y/q)/(p2+q2) 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∣+z=3+i (where i=−1​). Then ∣z∣ 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 the imaginary part of z is non-zero and a=z2+z+1 is real. Then a cannot take the value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all complex numbers z satisfying ∣z2+z+1∣=1. Then which of the following statements is/are TRUE?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for z=α+iβ, ∣z+2∣=z+4(1+i), then α+β and αβ are the roots of the equation

Animated Solution for Mathematics - Complex Numbers: Let $z = x + i y$ be a non-zero complex number such that $z^{2} = i ∣ z ∣^{2}$ , where $i = - 1$ , then $z$ lies on the:

Defining the Components of $z$

Simplifying the $i^{2}$ Term

Testing Case 1: $y = x$

Testing Case 2: $y = - x$

$z \neq = 0$ is a complex number

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number}$ . Then which of the following is NOT correct?

Let $z$ be a complex number with non-zero imaginary part. If $\frac{2 + 3 z + 4 z ^{2}}{2 - 3 z + 4 z ^{2}}$ is a real number, then the value of $∣ z ∣^{2}$ is _____________.

If $z = x - i y$ and $z^{1/3} = p + i q$ , then $(x / p + y / q) / (p^{2} + q^{2})$ is equal to

Let $z$ be a complex number such that $∣ z ∣ + z = 3 + i$ (where $i = - 1$ ). Then $∣ z ∣$ is equal to:

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

Let $S$ be the set of all complex numbers $z$ satisfying $∣ z^{2} + z + 1∣ = 1$ . Then which of the following statements is/are TRUE?

If for $z = α + i β$ , $∣ z + 2∣ = z + 4 (1 + i)$ , then $α + β$ and $α β$ are the roots of the equation