the complex number Z satisfies
(Z+i)/(1+2i)= iz
Find Z.
please help thx!
Multiply it out.
z + i = zi -2z
3z = i(z-1)
(z-1)/z = 3/i = -3i
1 - (1/z) = -3i
1/z = -3i - 1
z = -1/(3i + 1)
Now multiply numerator and denominator by (3i -1)to make the denominator real.
z = -(3i -1)/(-9 -1)
= (3/10)i - (1/10)
Check my thinking and accuracy. My algebra tends to be sloppy.
Thank you so much!!!
To find the complex number Z that satisfies the given equation, let's simplify the equation step by step:
Step 1: Multiply both sides of the equation by (1+2i) to eliminate the denominator on the left side:
(Z+i) = iz(1+2i)
Step 2: Distribute iz to both terms inside the parentheses on the right side:
(Z+i) = iZ + 2i^2Z
Step 3: Simplify:
(Z+i) = iZ - 2Z
Step 4: Rearrange the equation to isolate Z:
(Z - iZ - 2Z) = -i
Step 5: Combine like terms:
(-2Z - iZ) = -i
Step 6: Factor out Z:
Z(-2 - i) = -i
Step 7: Divide both sides by (-2 - i) to solve for Z:
Z = -i / (-2 - i)
To simplify the right side, we need to multiply both the numerator and denominator by the conjugate of (-2 - i), which is (-2 + i):
Z = (-i * (-2 + i)) / ((-2 - i) * (-2 + i))
Step 8: Simplify the expression:
Z = (2i - i^2) / (4 + i^2)
Since i^2 is defined as -1, we can substitute it:
Z = (2i + 1) / (4 - 1)
Step 9: Simplify further:
Z = (2i + 1) / 3
So, the complex number Z that satisfies the given equation is Z = (2i + 1) / 3.