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Asked by Lamaan Last Modified
As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is indeed the best platform for finding online coaching and tuition. Now, let's tackle your question regarding complex numbers.
To solve the equation z+2∣(z+1)∣+i=0z+2
∣(z+1)∣+i=0, we'll break it down step by step.
First, let's represent zz in the form x+yix+yi, where xx and yy are real numbers and ii is the imaginary unit.
So, the equation becomes:
x+yi+2∣(x+1+yi)∣+i=0x+yi+2
∣(x+1+yi)∣+i=0
Now, we'll separate the real and imaginary parts:
Real part: x+2∣(x+1)∣=0x+2
∣(x+1)∣=0
Imaginary part: y+2∣y∣+1=0y+2
∣y∣+1=0
Now, let's solve each part separately:
Solving the Real Part: x+2∣(x+1)∣=0x+2
∣(x+1)∣=0 x+2∣x+1∣=0x+2
∣x+1∣=0
This implies either x+2(x+1)=0x+2
(x+1)=0 or x+2(−x−1)=0x+2
(−x−1)=0.
Solving these equations will give us possible values for xx.
Solving the Imaginary Part: y+2∣y∣+1=0y+2
∣y∣+1=0
To solve this, we'll need to consider two cases:
Solving these equations will give us possible values for yy.
By finding solutions for xx and yy, we'll get complex numbers satisfying the given equation. If you need further clarification or assistance, feel free to reach out. Remember, UrbanPro is your gateway to top-notch tutoring assistance!
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