How do I figure these 2 out.
Find standard form a +bi
(4 square root -9) ( -3 square root -5)
x^2 -3x+4 evaluated for x=1+i
For the second one i got 1+3i is this correct
4 (-9)^.5 * -3 (-5)^.5
-12 (45)^.5
-12 (3*3*5)^.5
-36 sqrt 5
(1+i)^2 -3(1+i) + 4
=
1 + 2 i - 1 - 3 - 3 i + 4
=
(1 -i)
Doesnt i^2 equal -1
To find the standard form (a + bi) of the product of two complex numbers, you can follow these steps:
Step 1: Simplify each complex number individually.
In this case, simplify (4√-9) and (-3√-5).
The square root of -9 can be written as √-1 * √9, where √-1 is denoted as "i" (the imaginary unit) and √9 is 3. Therefore, 4√-9 becomes 4i * 3, which equals 12i.
Similarly, the square root of -5 can be written as √-1 * √5. Thus, -3√-5 becomes -3i * √5.
Step 2: Multiply the simplified complex numbers.
Now, multiply 12i * -3i * √5.
The product of two imaginary units (i * i) is -1, so we have -36 * √5.
Step 3: Write the answer in standard form.
To express the result in standard form (a + bi), separate the real (a) and imaginary (bi) parts.
The real part is -36 * √5, and the imaginary part is 0 (since there is no plain imaginary part). Thus, the standard form is -36√5 + 0i, which simplifies to -36√5.
To evaluate the expression x^2 - 3x + 4 for x = 1 + i, follow these steps:
Step 1: Substitute the value of x into the expression.
Replace every instance of x in the expression x^2 - 3x + 4 with 1 + i.
(1 + i)^2 - 3(1 + i) + 4
Step 2: Simplify the expression using the given value of x.
Expand and simplify the expression:
(1 + 2i + i^2) - 3 - 3i + 4
1 + 2i - 1 - 3 - 3i + 4
Combine like terms:
(1 - 1 + 4) + (2i - 3i) - 3
4 - i - 3
1 - i
Therefore, when x = 1 + i, the expression x^2 - 3x + 4 evaluates to 1 - i.