Question

find the multiplication inverse of the following complex number = (2-3i)^2

This question is from chaoter 5 class 11
Plz solve it!!! ​

Answer 1

Answer:

-7i is the answer of this question

Answer 2

Step-by-step explanation:

Given :-

(2-3i)²

To find:-

Find the Multiplicative Inverse of (2-3i)²?

Solution:-

Given that (2-3i)²

It is in the form of (a-b)²

Where , a = 2 and b = 3i

We know that

(a-b)² = a²-2ab+b²

(2-3i)² = 2²-2(2)(3i)+(3i)²

=> (2-3i)² = 4-12i+9i²

We know that i² = -1

=> (2-3i)² = 4-12i+9(-1)

=> (2-3i)² = 4-12i-9

=> (2-3i)² = -5-12i

=> (2-3i)² = -(5+12i)

The Multiplicative Inverse of -(5+12i) is 1/-(5+12i)

=>-1/(5+12i) or

The Rationalising factor of 5+12i is 5-12i

On Rationalising the denominator then

=> (-1/5+12i)×[(5-12i)/(5-12i)]

=> [(5-12i)(-1)]/[(5+12i)(5-12i)]

=> -(5-12i)/[5²-(12i)²]

Since (a+b)(a-b) = a²-b²

Where , a = 5 and b = 12i

=> -(5-12i)/(25-144i²)

=> -(5-12i)/(25-144(-1))

=> -(5-12i)/(25+144)

=> -(5-12i)/169

=> (-5+12i)/169

Answer:-

The Multiplicative Inverse of (2-3i)² is

-1/(5+12i)or (-5+12i)/169

Used formulae:-

• (a-b)² = a²-2ab+b²
• (a+b)(a-b) = a²-b²
• The Rationalising factor of a+ib is a-ib
• i² = -1
• The product of two numbers is one then they are called Multiplicative Inverse of each other.
• a×(1/a) = 1