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Rotate and Stretch Complex Number 3+4i by π/4 | JEE

JEE Maths question with a full step-by-step solution.

Question

The complex number

3 + 4i

is rotated about the origin by an angle

\dfrac{\pi}{4}

in the anticlockwise direction and then stretched twice. The complex number corresponding to the new position is

\sqrt{2}(-3 + 4i)

\sqrt{2}(-1 + 7i)

correct

\sqrt{2}(3 - 4i)

\sqrt{2}(-1 - 7i)

Solution

Step 1: Recall the transformation rules. Rotating a complex number

z

anticlockwise about the origin by angle

\phi

multiplies it by

e^{i\phi}

. Stretching (scaling) the length by a factor

k

multiplies it by

k

. Here

\phi = \dfrac{\pi}{4}

and

k = 2

, so the new number is:

w = 2\,e^{i\pi/4}\,(3 + 4i)

Step 2: Write

e^{i\pi/4}

in rectangular form:

e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \frac{1}{\sqrt 2} + \frac{i}{\sqrt 2} = \frac{1 + i}{\sqrt 2}

Step 3: Substitute:

w = 2\cdot\frac{1 + i}{\sqrt 2}\,(3 + 4i) = \frac{2}{\sqrt 2}(1 + i)(3 + 4i) = \sqrt{2}\,(1 + i)(3 + 4i)

Step 4: Expand the product

(1 + i)(3 + 4i)

(1 + i)(3 + 4i) = 3 + 4i + 3i + 4i^2 = 3 + 7i - 4 = -1 + 7i

Step 5: Therefore:

w = \sqrt{2}\,(-1 + 7i)

Correct answer: (2)

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