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Question

Question: Let z be a complex number such that \(\left| z \right|+z=3+i\)(where \(i=\sqrt{-1}\)). Then \(\left|...

Let z be a complex number such that z+z=3+i\left| z \right|+z=3+i(where i=1i=\sqrt{-1}). Then z\left| z \right| is equal to:-
A. 54\dfrac{5}{4}
B. 414\dfrac{\sqrt{41}}{4}
C. 343\dfrac{\sqrt{34}}{3}
D. 53\dfrac{5}{3}

Explanation

Solution

We must assume z as a general complex number, z=a+ibz=a+ib then the magnitude of the complex number zz will be z=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}using these two relations we must substitute them in the equation. By comparing real and imaginary parts we get value of aa and bb.And finally we use these values to find z\left| z \right|.

Complete step by step answer:
A complex number is the one which cannot be represented on a real number line alone so we also include a imaginary number line to collectively represent the number.
A complex number has two parts real part and a imaginary part as z=a+ibz=a+ib a,bRa,b\in \mathbb{R} (where i=1i=\sqrt{-1})
Where aa is called real part and ibib is complex part,
We have been given the following equation in the question,
z+z=3+i\left| z \right|+z=3+i
We assume zz as any general complex number to facilitate easy solving of question
Let z=a+ibz=a+ib,
The modulus or absolute value of any complex number is given by –
z=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}
Having all the necessary informations we will now substitute them in the given equation,
a2+b2+a+ib=3+i\sqrt{{{a}^{2}}+{{b}^{2}}}+a+ib=3+i
We will shift a+iba+ib to the right side,
a2+b2=3+i(a+ib)\sqrt{{{a}^{2}}+{{b}^{2}}}=3+i-(a+ib)
Since we have only real number on left hand, we can write complex coefficient as 0.
a2+b2+0i=(3a)+i(1b)\sqrt{{{a}^{2}}+{{b}^{2}}}+0i=(3-a)+i(1-b)
We compare the real and imaginary parts,
First we are comparing imaginary part,
0i=i(1b)0i=i(1-b)
b=1b=1
Now we equate real parts,
a2+b2=3a\sqrt{{{a}^{2}}+{{b}^{2}}}=3-a
Using b=1b=1
a2+b2=3a\sqrt{{{a}^{2}}+{{b}^{2}}}=3-a
Squaring both sides,
a2+1=a2+96a{{a}^{2}}+1={{a}^{2}}+9-6a
a=43a=\dfrac{4}{3}
Since z=a2+b2\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}} we write,
z=1+(43)2\left| z \right|=\sqrt{1+{{\left( \dfrac{4}{3} \right)}^{2}}}
z=1+169\left| z \right|=\sqrt{1+\dfrac{16}{9}}
z=53\left| z \right|=\dfrac{5}{3}

So, the correct answer is “Option D”.

Note: We must take care that substitution is done properly, when comparing real and imaginary parts be sure to compare real part with real part and imaginary part with imaginary part. Students can also write a+iba+ib as x+iyx+iy since change of variable doesn’t affect the number on the complex graph.