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

& {{z}_{1+1}}=z_{1}^{2}+i \\\ & {{z}_{2}}=0+i \\\ \end{aligned}$$ Hence, $${{z}_{2}}=i$$ When n = 2, $${{z}_{2+1}}={{\left( {{z}_{2}} \right)}^{2}}+i$$ {$$\because $$ We know that, $${{i}^{2}}=-1$$} $$\begin{aligned} & {{z}_{2+1}}={{\left( i \right)}^{2}}+i=-1+i \\\ & \therefore {{z}_{3}}=-1+i \\\ \end{aligned}$$ When n = 3, $${{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{aligned} & \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{aligned}$$ Hence, $${{z}_{4}}=-i$$ When n = 4, $${{z}_{4+1}}={{\left( {{z}_{4}} \right)}^{2}}+i$$ $$\begin{aligned} & {{z}_{5}}={{\left( -i \right)}^{2}}+i={{1}^{2}}+i \\\ & \therefore {{z}_{5}}=-1+i \\\ \end{aligned}$$ Thus, $${{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{aligned} & {{z}_{6}}={{\left( -1+i \right)}^{2}}+i \\\ & {{z}_{6}}=-i \\\ \end{aligned}$$ Where, $${{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.