1. z ≠ 0 is a complex number Column I...

MathematicsAlgebraic Operations on Complex NumbersJEE Advanced 1992Easy

$z \neq = 0$ is a complex number

List-I

(P)

R ez = 0

(Q)

A r g z = \frac{π}{4}

List-II

(1)

R e z^{2} = 0

(2)

I m z^{2} = 0

(3)

R e z^{2} = I m z^{2}

Answer:P → 2, Q → 1

Visualized Solution (Hindi)

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

Let the complex number be $z = x + i y$
Here, $x = R e (z)$ and $y = I m (z)$
We are given $z \neq = 0$ , so $x$ and $y$ are not both zero.

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z^{2}

Calculate $z^{2} = (x + i y)^{2}$
Using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ :
$z^{2} = x^{2} + 2 i x y + (i y)^{2}$
Since $i^{2} = - 1$ , we get:
$z^{2} = (x^{2} - y^{2}) + i (2 x y)$

Condition (A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R e (z) = 0

Condition (A) states $R e (z) = 0$
This implies $x = 0$
The complex number simplifies to $z = i y$

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x = 0 into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z^{2}

Substitute $x = 0$ into $z^{2} = (x^{2} - y^{2}) + i (2 x y)$
$z^{2} = (0^{2} - y^{2}) + i (2 \cdot 0 \cdot y)$
$z^{2} = - y^{2} + i (0)$

Result for (A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I m (z^{2}) = 0

Since $z^{2} = - y^{2}$ is a purely real number:
$I m (z^{2}) = 0$
Conclusion for (A): Matches with (q)

Condition (B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A r g (z) = \frac{π}{4}

Condition (B) states $A r g (z) = \frac{π}{4}$
This implies $tan (\frac{π}{4}) = \frac{y}{x} = 1$
Therefore, $x = y$ (and $x > 0$ for the first quadrant)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x = y into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z^{2}

Substitute $y = x$ into $z^{2} = (x^{2} - y^{2}) + i (2 x y)$
$z^{2} = (x^{2} - x^{2}) + i (2 \cdot x \cdot x)$
$z^{2} = 0 + i (2 x^{2})$

Result for (B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R e (z^{2}) = 0

Since $z^{2} = i (2 x^{2})$ is a purely imaginary number:
$R e (z^{2}) = 0$
Conclusion for (B): Matches with (p)

Final Summary and Takeaway

Final Match:
(A) $R e (z) = 0 ⟹ (q) I m (z^{2}) = 0$
(B) $A r g (z) = \frac{π}{4} ⟹ (p) R e (z^{2}) = 0$
Key Takeaway: Squaring a complex number squares its magnitude and doubles its argument.

00:00 / 00:00

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(A)

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Answer:C

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If $z$ and $ω$ are two non-zero complex numbers such that $∣ z ω ∣ = 1$ and $A r g (z) - A r g (ω) = π /2$ , then $\overset{z}{ˉ} ω$ is equal to

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Answer:C

$z \neq = 0$ is a complex number

List-I

List-II

Visualized Solution (Hindi)

Defining $z = x + i y$

Expanding $z^{2}$

Condition (A): $R e (z) = 0$

Substituting $x = 0$ into $z^{2}$

Result for (A): $I m (z^{2}) = 0$

Condition (B): $A r g (z) = \frac{π}{4}$

Substituting $x = y$ into $z^{2}$

Result for (B): $R e (z^{2}) = 0$

Final Summary and Takeaway

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$z \neq = 0$ is a complex number

List-I

List-II

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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

.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

Visualized Solution (Hindi)

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=x+iy

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

Condition (A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Re(z)=0

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2

Result for (A): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Im(z2)=0

Condition (B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Arg(z)=4π​

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=y into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2

Result for (B): .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Re(z2)=0

Final Summary and Takeaway

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List-I

List-II

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$z \neq = 0$ is a complex number

Defining $z = x + i y$

Expanding $z^{2}$

Condition (A): $R e (z) = 0$

Substituting $x = 0$ into $z^{2}$

Result for (A): $I m (z^{2}) = 0$

Condition (B): $A r g (z) = \frac{π}{4}$

Substituting $x = y$ into $z^{2}$

Result for (B): $R e (z^{2}) = 0$

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If $z = (\frac{3}{2} + \frac{i}{2})^{5} + (\frac{3}{2} - \frac{i}{2})^{5}$ , then

Let $z$ and $w$ be complex numbers such that $\overset{z}{ˉ} + i \overset{w}{ˉ} = 0$ and $a r g z w = π$ . Then $a r g z$ equals

$z$ and $w$ are two nonzero complex numbers such that $∣ z ∣ = ∣ w ∣$ and $A r g z + A r g w = π$ then $z$ equals

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If $z$ is a complex number of unit modulus and argument $θ$ , then $a r g (\frac{1 + z}{1 + z ˉ})$ equals

If $a r g (z) < 0$ , then $a r g (- z) - a r g (z) =$

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

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