If $z_1 = -1$, $z_2 = -i$ then find $\mathrm{Arg}(z_1 . z_2)$.

If z₁ = -1, z₂ = -i then find mathrmArg(z₁ . z₂).

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z_1 = -1

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 2mImportance★★★★★

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Concept understanding — Complex Number Transform

The Intuition: Why "Transform" a Number?

You already know that a real number lives on a line — the number line. Adding $+2$ slides you right; multiplying by $-1$ flips you to the opposite side. But what if you want to rotate something? A real number can't do that on its own. Multiplying by $-1$ is a 180° rotation, but what about a 90° rotation? That's where the complex number transform comes in.

Think of a complex number $z = a + bi$ as a point (or an arrow) on a 2D plane. The real part $a$ is the horizontal coordinate, the imaginary part $b$ is the vertical coordinate. Now, when you multiply two complex numbers, something beautiful happens: the lengths multiply, and the angles add.

Note

This is the core insight: multiplication of complex numbers is a rotation + scaling operation, not just a scaling like real numbers.

So a "complex number transform" is simply the act of applying a complex number (as an operator) to another complex number (as a point) — usually by multiplication — to achieve a geometric transformation: rotation, scaling, or both.

The Precise Statement

Let $z = x + yi$ be any complex number (the "point" you want to transform).

Let $w = r(\cos\theta + i\sin\theta)$ be a fixed complex number (the "transformer").

Then the complex number transform of $z$ by $w$ is:

w \cdot z = r(\cos\theta + i\sin\theta) \cdot (x + yi)

When you multiply this out (using $i^2 = -1$ ), the result is a new complex number $z'$ whose geometric meaning is:

Scale the distance of $z$ from the origin by a factor of $r$
Rotate the point $z$ around the origin by an angle $\theta$ counterclockwise

\text{If } w = re^{i\theta}, \text{ then } w \cdot z \text{ rotates } z \text{ by } \theta \text{ and scales it by } r.

This is often written using Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$ , so $w = re^{i\theta}$ .

A Concrete Example

Take the point $z = 1 + 0i$ (the number 1 on the real axis).

Let the transformer be $w = i$ (which has $r=1$ , $\theta = 90^\circ$ ).

i \cdot 1 = i

The point $(1,0)$ moved to $(0,1)$ — a 90° rotation counterclockwise. No scaling because $|i| = 1$ .

Now take $z = 2 + 0i$ and $w = 2i$ (which has $r=2$ , $\theta = 90^\circ$ ):

2i \cdot 2 = 4i

The point $(2,0)$ moved to $(0,4)$ — rotated 90° and scaled by factor 2.

Watch out

A common mistake: thinking that multiplying by $i$ always gives a 90° rotation. It does — but only if you multiply the entire complex number. Multiplying just the real part by $i$ is not the same as multiplying the whole number.

Why This Matters

This transform is the foundation of: …

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Multiply $z_1$ and $z_2$ first, then find the argument of the product.

Given $z_1=-1$ , $z_2=-i$ .

$z_1 z_2 = (-1)(-i) = i$

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Z = (-\sqrt{3} - i)$
(a) $2\left(\cos\dfrac{5\pi}{6} - i\sin\dfrac{5\pi}{6}\right)$
(b) $2\left(-\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right)$
(c) $2\left(\cos\dfrac{5\pi}{6} + i\sin\dfrac{5\pi}{6}\right)$
(d) None of these
›Reveal solutionSolution

$Z=-\sqrt3 - i$ has modulus $2$ and lies in the third quadrant, giving principal argument $-5\pi/6$ ; writing $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$ produces option (a).

$Z = -\sqrt{3} - i, \qquad a = -\sqrt3,\ b=-1$

Modulus:

$r = \sqrt{a^2+b^2} = \sqrt{3+1} = 2$

Both $a$ and $b$ are negative, so $Z$ lies in the third quadrant. The reference angle is:

$\tan\alpha = \left|\frac{b}{a}\right| = \frac{1}{\sqrt3} \implies \alpha = \frac{\pi}{6}$

For a third-quadrant point, the principal argument (in $(-\pi,\pi]$ ) is:

$\theta = -\left(\pi - \frac{\pi}{6}\right) = -\frac{5\pi}{6}$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The polar form of a complex number is$
(a) $\sin\theta + i\cos\theta$
(b) $\cos\theta + i\sin\theta$
(c) $r(\cos\theta + i\sin\theta)$
(d) $r(\sin\theta + i\cos\theta)$
›Reveal solutionSolution

The polar form expresses a complex number using its modulus $r$ and argument $\theta$ : $r(\cos\theta+i\sin\theta)$ .

Any complex number $z=x+iy$ can be written using its modulus $r=\sqrt{x^2+y^2}$ and argument $\theta$ (the angle it makes with the positive real axis), since $x=r\cos\theta$ and $y=r\sin\theta$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$Z=-\sqrt3+i$
(a) $\dfrac{\pi}{6}$
(b) $\dfrac{5\pi}{6}$
(c) $-\dfrac{\pi}{6}$
(d) $-\dfrac{5\pi}{6}$
›Reveal solutionSolution

The argument of $Z=-\sqrt3+i$ is $\dfrac{5\pi}{6}$ .

Here $Z=x+iy$ with $x=-\sqrt3$ (negative) and $y=1$ (positive), so $Z$ lies in the second quadrant.

The reference angle $\alpha$ satisfies $\tan\alpha = \dfrac{|y|}{|x|} = \dfrac{1}{\sqrt3}$ , so $\alpha=\dfrac{\pi}{6}$ .

…

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CBSE 2020Set ANNUAL1 mark
$1+i$
›Reveal solutionSolution

For $1+i$ , modulus $=\sqrt{2}$ and argument $=\frac{\pi}{4}$ , so the statement is True.

For a complex number $x+iy$ , modulus $r=\sqrt{x^2+y^2}$ and argument $\theta=\tan^{-1}(y/x)$ .

Here $x=1$ , $y=1$ :

$r = \sqrt{1^2+1^2} = \sqrt{2}$

Since both $x>0$ and $y>0$ , the point lies in the first quadrant, so:

$\theta = \tan^{-1}(1) = \frac{\pi}{4}$

…

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←Previous chapterCh 3Differential Equations

Q.If z1=−1z_1 = -1z1​=−1, z2=−iz_2 = -iz2​=−i then find Arg(z1.z2)\mathrm{Arg}(z_1 . z_2)Arg(z1​.z2​).

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Q.If $z_1 = -1$ , $z_2 = -i$ then find $\mathrm{Arg}(z_1 . z_2)$ .

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