Question

ii) (1+i)(1-1)
i is iota find the value of a and b by coverting in a+bi ...
2 one plxz​

Answer 1

Answer:

that's your answer, for more follow me, I am also in class XIth

Answer 2

Answer

1 ) a = -4 b = -3

2). a =. 0 b = 1

Case 1st

→ (1+2i) (-2+i ) = Z

Multiplying these two complex numbers.

-2(1+2i) +i(1+2i)

-2 -4i +i +2i² = Z

✪We know that i² = -1, so

-2 - 4i + i + 2(-1) = Z

-2 - 4i + i - 2. = Z

-2 -2 -4i + i. = Z

-4 -3i. = Z. ( 1)

We know that the standard form of Z is a+ ib. (2)

Comparing Z(Re) with Z(Re) and Z(Im) with Z (Im)

We got

-4 = a

-3 = b.

Case 2nd

→ (1+i) (1-i )-¹ = Z

As we know that i-¹ = 1/i , so

$(1 + i) \: ( \frac{1}{1 - i})$

$\frac{1 + i}{1 - i}$

Now we have to rationalize the above fraction.

$\frac{(1 + i)}{(1 - i)} \times \frac{(1 + i)}{(1 + i)}$

(a+b)(a-b) = a² -b²

$\frac{( {1 + i)}^{2} }{( {1)}^{2} - {i}^{2} }$

As we know that i² = -1 applying this

$\frac{( {1 + i)}^{2} }{1 - ( - 1)} = \frac{( {1)}^{2} + ( {i)}^{2} + 2i }{1 + 1}$

(a+b)² = a² + b² +2ab

$\frac{2i}{2} = i$

Here 2 is cancelled out.

So we got Z as

Z = 0 + 1i

Comparing it with general form of Z

a + ib = 0 + 1i

a = 0

b = 1

Hope it helps