Given:
The number is 
.
To find:
The a+bi form of given number.
Solution:
We have,
It can be written as
![[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%7Bab%7D%3D%5Csqrt%7Ba%7D%5Csqrt%7Bb%7D%5D)
![[\because \sqrt{-1}=i]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%7B-1%7D%3Di%5D)
Here, real part is missing. So, it can be taken as 0.
So, a = 0 and 
.
Therefore, the a+bi form of given number is 
.
Answer:
<em>-21 and 4</em>
Step-by-step explanation:
They multiply to a negative number, so you need a positive number and a negative number.
In which language?
Please tell me the language java, c++, cetc?
9514 1404 393
Answer:
1a: x+3 = 5
1c: 6 = 2z
2b: x = 2
2d: 3 = z
3: the solutions make the hangars balance
Step-by-step explanation:
1. We can write the equations by listing the contents of the hangar and using an equal sign to show the balance between left side and right side. It can work well to put left side contents of the hangar on the left side of the equal sign.
A: x + 3 = 5
C: 1 + 1 + 1 + 1 + 1 + 1 = z + z simplifies to 6 = 2z
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2. B: We can subtract 3 from both sides of the hangar (and equation) to find the value of x.
(x +3) -3 = 5 -3
x = 2 . . . . . hangar balances with 2 on the right
D: We can divide both sides of the hangar by 2, splitting the content into two equal parts. Then one of those parts can be removed from each side.
2(3) = 2(z)
3 = z . . . . . . hangar balances with 3 on the left
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3. The found values will keep the hangar in balance when they are substituted for the corresponding variables.
A: 2 + 3 = 5
C: 1 + 1 + 1 + 1 + 1 + 1 = 3 + 3
