Question
Question: If magnitude of a complex number \[z=4-3i\] is tripled and is rotated anti clockwise by an angle of ...
If magnitude of a complex number z=4−3i is tripled and is rotated anti clockwise by an angle of π , then the resulting complex number would be
1. −12+9i
2. 12+9i
3. 7−6i
4. 7+6i
Solution
To solve this question you must first start by taking the magnitude of the given complex number. Let the complex number be z and its magnitude will be ∣z∣ and the formula to find its magnitude will be a2+b2 . Now we need to find the complex number which we will obtain after rotating the complex number z. Let the resulting complex number be z1 and we can find it by using the formula z1=e−iπ
Complete step by step answer:
Now to start the solution let us assume that
z=4−3i
Now to find the magnitude of z we can use it by the formula that
∣z∣=a2+b2
Here we know that a is the real part of the complex number and b is the imaginary part of it. Therefore substituting their values inside the formula we get
∣z∣=42+32
Simplifying we get
∣z∣=5
Now we also know that the complex number is tripled in size therefore the magnitude of required complex number will be
∣z1∣=5×3
∣z1∣=15
Now that we know the magnitude we know that the complex number is then rotated anticlockwise by π which means that the new complex number vector will be
z1=e−iπz=(cosiπ−siniπ)z
Therefore simplifying we get
z1=−4+3i
Now we know that the new vector is in the direction of the unit vector which is −54+53i
Now henceforth our required vector we can say will be since its magnitude is 15
z1=15(−54+53i)
z1=−12+9i
So, the correct answer is “Option 1”.
Note: Argand plane is a plane where we can represent any complex number in the easiest way. Here in this plane the x axis which is the horizontal axis represents all real numbers whereas the y axis which is the vertical axis represents all the imaginary numbers .