Question

Let z=6-7i and z^2=a+bi. What are the values of a and b ?

Answer 1

Answer:

The values are

a = -13

b = -84

Step-by-step explanation:

Given:

z=6-7i and

$z^{2}=a + bi$

To Find:

a = ?

b = ?

Solution:

Z is Complex Number consist of Real part and Imaginary part

We have

$z=6-7i\\\textrm{squaring on both the side we get}\\z^{2}=(6-7i)^{2}$

Using the identity $(A-B)^{2} =A^{2}-2AB+ B^{2}$ we get

$z^{2} =6^{2}-2\times 6\times 7i+ (7i)^{2}\\z^{2} =36-84i+49i^{2}$

i² = -1

∴ $z^{2} =36-84i+49(-1)\\z^{2} =36-49-84i\\z^{2} =-13-84i$

Now on comparing with  $z^{2}=a + bi$  equation we get

a = -13

b = -84