The point represented by the complex number $2-i$ is rotated about origin through an angle $\dfrac{\pi}{2}$ in the clockwise direction, the new position of point is: (A) $1+2i$ (B) $-1-2i$ (C) $2+i$ (D) $-1+2i$

The point represented by the complex number 2-i is rotated about origin through an angle (π)/(2) in the clockw

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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Rotating a point $90\degree$ clockwise about the origin corresponds to multiplying its complex number by $-i$ . For $z=2-i$ , this gives $z(-i) = -1-2i$ — option (B).

Why multiplying by $-i$ means a $90\degree$ clockwise turn

Look at what multiplying by $i$ does to a general point $w=x+iy$ :

iw = i(x+iy) = ix+i^2y = -y+ix

So $w=(x,y)$ becomes the point $(-y,x)$ . Checking a few examples confirms this is a $90\degree$ counter-clockwise turn about the origin — e.g. $1=(1,0)$ becomes $i\cdot1=i=(0,1)$ , and $i=(0,1)$ becomes $i\cdot i=i^2=-1=(-1,0)$ : each multiplication by $i$ swings the point one quarter-turn counter-clockwise.

Multiplying by $-i$ is the reverse operation — since $i\cdot(-i)=-i^2=1$ , multiplying by $-i$ exactly undoes a multiplication by $i$ — so it must be a $90\degree$ clockwise turn. Checking algebraically for $w=x+iy$ :

-iw = -i(x+iy) = -ix-i^2y = y-ix

So $w=(x,y)$ becomes $(y,-x)$ — the coordinate rule for a clockwise rotation.

Applying this to the problem

Identify the point. $z=2-i$ corresponds to $(x,y)=(2,-1)$ .
The rotation is $90\degree$ clockwise, so multiply $z$ by $-i$ :

$z' = z\cdot(-i) = (2-i)(-i)$ …

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Q.The point represented by the complex number 2−i2-i2−i is rotated about origin through an angle π2\dfrac{\pi}{2}2π​ in the clockwise direction, the new position of point is: (A) 1+2i1+2i1+2i (B) −1−2i-1-2i−1−2i (C) 2+i2+i2+i (D) −1+2i-1+2i−1+2i

Why multiplying by −i-i−i means a 90°90\degree90° clockwise turn

Applying this to the problem

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Q.The point represented by the complex number $2-i$ is rotated about origin through an angle $\dfrac{\pi}{2}$ in the clockwise direction, the new position of point is:
(A) $1+2i$
(B) $-1-2i$
(C) $2+i$
(D) $-1+2i$

Why multiplying by $-i$ means a $90\degree$ clockwise turn

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