Question

Let \[i=\sqrt{-1}\], define a sequence of complex number by \[{{z}_{1}}=0,{{z}_{n+1}}=z_{n}^{2}+i\] for \[n\ge 1\]. In the complex plane, then \[{{z}_{111}}\] lies in which quadrant?
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

Hint: It is said that \[n\ge 1\], thus put, n = 1, 2, 3….. in the expression \[{{z}_{n+1}}=z_{n}^{2}+i\]. Thus find values of \[{{z}_{1}},{{z}_{2}},{{z}_{3}}.....,{{z}_{10}}\]. Now compare these values to get a sequence. Thus find \[{{z}_{111}}\] and determine the quadrant by taking its real and imaginary part.

Complete step-by-step answer:
We have been given the sequence of complex number by \[{{z}_{1}}=0\]and \[{{z}_{n+1}}=z_{n}^{2}+i\]. We need to find where \[{{z}_{111}}\] lies on the quadrant and given that \[n\ge 1\]. So, n = 1, 2, 3….
Now, \[{{z}_{1}}=0\] and \[{{z}_{n+1}}=z_{n}^{2}+i\]
When n = 1, \[{{z}_{n+1}}=z_{n}^{2}+i\] becomes
\[\begin{align}
  & {{z}_{1+1}}=z_{1}^{2}+i \\
 & {{z}_{2}}=0+i \\
\end{align}\]
Hence, \[{{z}_{2}}=i\]
When n = 2, \[{{z}_{2+1}}={{\left( {{z}_{2}} \right)}^{2}}+i\] {\[\because \] We know that, \[{{i}^{2}}=-1\]}
\[\begin{align}
  & {{z}_{2+1}}={{\left( i \right)}^{2}}+i=-1+i \\
 & \therefore {{z}_{3}}=-1+i \\
\end{align}\]
When n = 3, \[{{z}_{3+1}}={{\left( {{z}_{3}} \right)}^{2}}+i\]
\[{{z}_{4}}={{\left( -1+i \right)}^{2}}+i\]
\[{{\left( -1+i \right)}^{2}}\] is of the form, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\].
\[\begin{align}
  & \therefore {{z}_{4}}={{\left( -1 \right)}^{2}}-2\times 1\times i+{{i}^{2}}+i \\
 & {{z}_{4}}=1-2i+{{i}^{2}}+i \\
 & {{z}_{4}}=1-1-i=-i \\
\end{align}\]
Hence, \[{{z}_{4}}=-i\]
When n = 4, \[{{z}_{4+1}}={{\left( {{z}_{4}} \right)}^{2}}+i\]
\[\begin{align}
  & {{z}_{5}}={{\left( -i \right)}^{2}}+i={{1}^{2}}+i \\
 & \therefore {{z}_{5}}=-1+i \\
\end{align}\]
Thus, \[{{z}_{5}}\] is equal to \[{{z}_{3}}\] i.e. \[{{z}_{5}}={{z}_{3}}=-1+i\].
When n = 5, \[{{z}_{5+1}}={{\left( {{z}_{5}} \right)}^{2}}+i\]
\[\begin{align}
  & {{z}_{6}}={{\left( -1+i \right)}^{2}}+i \\
 & {{z}_{6}}=-i \\
\end{align}\]
Where, \[{{z}_{6}}={{z}_{4}}=-i\].
Hence taking the summary of what we found above,
\[{{z}_{1}}=0,{{z}_{4}}=-i,{{z}_{7}}=-1+i\], when n = 6.
\[{{z}_{2}}=i,{{z}_{5}}=-1+i,{{z}_{8}}=-i\], when n = 7.
\[{{z}_{3}}=-1+i,{{z}_{6}}=-i,{{z}_{9}}=-1+i\], when n = 8.
Thus we get that, \[{{z}_{3}}={{z}_{5}}=-1+i\] and \[{{z}_{4}}={{z}_{6}}=-i\].
Hence, we get \[{{z}_{7}}={{z}_{5}}=-1+i\]. Similarly, \[{{z}_{8}}=-i\], equal to \[{{z}_{6}}\].
Thus, \[{{z}_{3}}={{z}_{5}}={{z}_{7}}={{z}_{9}}={{z}_{11}}=-1+i\].
Similarly, \[{{z}_{4}}={{z}_{6}}={{z}_{8}}={{z}_{10}}=-i\].
Thus from this we can say that the odd terms are becoming equal to \[\left( -1+i \right)\] and the even value is becoming equal to \[\left( -i \right)\]. Hence, \[{{z}_{111}}\] is an odd term, we can find its value as,
\[{{z}_{111}}=z_{110}^{2}+i={{z}_{109}}\]
Now the value of \[{{z}_{111}}\] will be equal to value of \[{{z}_{109}}\], which in turn will be equal to \[{{z}_{107}}\] and thus it will go on until \[{{z}_{3}}\].
Hence, \[{{z}_{111}}={{z}_{109}}={{z}_{107}}={{z}_{105}}={{z}_{103}}=......={{z}_{7}}={{z}_{5}}={{z}_{3}}\].
Hence we can say that, \[{{z}_{111}}=-1+i\].
Now we need to find the quadrant in which, \[{{z}_{111}}=-1+i\] lies in. Here the real part is negative and the imaginary part is positive. Hence this lies in the \[{{2}^{nd}}\] quadrant.

Hence this lies in the \[{{2}^{nd}}\] quadrant. Hence, \[\left( -1+i \right)\] can be marked as (-1, 1) in the \[{{2}^{nd}}\] quadrant as shown in the figure.

Thus we got that \[{{z}_{111}}\] lies in \[{{2}^{nd}}\] quadrant.

\[\therefore \] Option (b) is the correct answer.

Note: The basic concept is that you substitute n = 1, 2, 3 …… in \[{{z}_{n+1}}=z_{n}^{2}+i\]. You should be careful while applying and remember, \[{{i}^{2}}=\left( -1 \right)\], which is one of the basics of complex numbers. To find the quadrant, compare the real and imaginary part. It its (1, 1) the \[{{1}^{st}}\], (-1, 1) then \[{{2}^{nd}}\], (-1, -1) in \[{{3}^{rd}}\] and (+1, -1) in \[{{4}^{th}}\] quadrant.